- Focus on the denominator of the exponent. - What root index corresponds to that denominator?

1. An exponent of \(\frac{1}{n}\) means an \(n\)th root. 2. Therefore, \(81^{\frac{1}{4}}=\sqrt[4]{81}\).

\(\sqrt[4]{81}\)

- The denominator of a rational exponent controls the root index. - Consider whether an odd-index root allows negative radicands.

1. The denominator \(5\) gives the root index, and the numerator \(3\) gives the power. 2. Thus, \(x^{\frac{3}{5}}=\sqrt[5]{x^3}\). 3. A fifth root is defined for every real radicand, so the real domain is all real numbers.

\(\sqrt[5]{x^3}\); domain: \(\mathbb{R}\)

- Use the radical index as the denominator of the exponent. - Use the power inside the radical as the numerator.

1. A fourth root corresponds to an exponent with denominator \(4\). 2. The power \(3\) in the radicand becomes the numerator. 3. Therefore, \(\sqrt[4]{y^3}=y^{\frac{3}{4}}\).

\(y^{\frac{3}{4}}\)

- Deal with the negative sign in the exponent before interpreting the fraction. - A negative exponent changes a power into its reciprocal.

1. A negative exponent means take the reciprocal of the corresponding positive power. 2. Therefore, \(16^{-\frac{3}{4}}=\frac{1}{16^{\frac{3}{4}}}\).

\(\frac{1}{16^{\frac{3}{4}}}\)

- Interpret each exponent as a radical first. - Compare the real-domain rule for even-index and odd-index roots.

1. The exponent \(\frac{1}{2}\) represents a square root, so its real input must satisfy \(x\ge0\). 2. The exponent \(\frac{1}{3}\) represents a cube root, which is defined for every real input.

\(x^{\frac{1}{2}}\) has domain \([0,\infty)\). \(x^{\frac{1}{3}}\) has domain \(\mathbb{R}\).

- What does the square-root symbol mean when the number under it is positive? - Does the negative sign in each expression get squared or stay outside the radical?

1. Since \((-5)^2=25\), the principal square root is \(\sqrt{25}=5\). 2. The minus sign in \(-\sqrt{25}\) is outside the radical, so the result is \(-5\). 3. The square-root symbol names the nonnegative square root; the outside negative sign changes its sign.

a) \(5\) b) \(-5\) The principal square root is nonnegative. In b), the negative sign is outside the square root.

- Check what number lies under each root at the stated input. - Can a negative real number have an odd root? An even root?

1. At \(x=0\), the square-root expression becomes \(\sqrt{-1}\), which is not real. 2. The cube-root expression becomes \(\sqrt[3]{-1}=-1\), which is real.

Only \(\sqrt[3]{x-1}\) has a real value at \(x=0\); its value is \(-1\).

- Look for perfect squares just below and above the radicand. - Taking positive square roots keeps the order of nonnegative numbers.

1. The consecutive perfect squares around \(13\) are \(9=3^2\) and \(16=4^2\). 2. Therefore, \(3<\sqrt{13}<4\).

Between \(3\) and \(4\), since \(3^2=9<13<16=4^2\).

- How is the area of a square related to its side length? - How does the principal square root differ from the two solutions of a squared equation?

1. The area of a square is the side length squared, so \(s^2=49\). 2. The principal square root is nonnegative: \(s=\sqrt{49}=7\,\text{in.}\). 3. Although \(-7\) also has square \(49\), a physical length cannot be negative.

\(s^2=49\); \(s=7\,\text{in.}\). A negative side length is not possible.

- Think about which power undoes cubing. - How does a root reverse raising a number to a power?

1. In the rational exponent \(\frac{1}{3}\), the denominator \(3\) identifies the root index. 2. Therefore, \(a^{\frac{1}{3}}=\sqrt[3]{a}\) for real \(a\).

\(\sqrt[3]{a}\). The denominator \(3\) indicates a cube root.

- Use the product and quotient properties of square roots when appropriate. - Look for perfect squares in the numerator, denominator, or radicand. - Check a decimal square root by squaring your result.

1. \(\sqrt{\frac{144}{25}}=\frac{\sqrt{144}}{\sqrt{25}}=\frac{12}{5}=2.4\). 2. \(\sqrt{2}\cdot\sqrt{18}=\sqrt{36}=6\). 3. Since \(0.02^2=0.0004\), \(\sqrt{0.0004}=0.02\). 4. \(\sqrt{10^4}=\sqrt{10{,}000}=100\).

a) \(2.4\) b) \(6\) c) \(0.02\) d) \(100\)

- Convert each mixed number to an improper fraction. - Use \(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\) for positive \(a\) and \(b\). - Identify perfect squares in each numerator and denominator.

1. Rewrite the mixed numbers: \(1\frac{11}{25}=\frac{36}{25}\), \(1\frac{13}{36}=\frac{49}{36}\), \(1\frac{15}{49}=\frac{64}{49}\), and \(1\frac{19}{81}=\frac{100}{81}\). 2. For a), \(\sqrt{\frac{36}{25}}=\frac{6}{5}\). 3. For b), \(\sqrt{\frac{49}{36}}=\frac{7}{6}\). 4. For c), \(\sqrt{\frac{64}{49}}=\frac{8}{7}\). 5. For d), \(\sqrt{\frac{100}{81}}=\frac{10}{9}\).

a) \(\frac{6}{5}\) b) \(\frac{7}{6}\) c) \(\frac{8}{7}\) d) \(\frac{10}{9}\)

- Recognize perfect-square whole numbers and decimals. - Combine a quotient of square roots under one radical. - Convert the mixed number to an improper fraction.

1. \(\sqrt{225}\cdot\sqrt{0.04}=15\cdot0.2=3\). 2. \(\frac{\sqrt{1000}}{\sqrt{10}}=\sqrt{\frac{1000}{10}}=\sqrt{100}=10\). 3. Since \(1\frac{7}{9}=\frac{16}{9}\), \(\sqrt{1\frac{7}{9}}=\sqrt{\frac{16}{9}}=\frac{4}{3}\).

a) \(3\) b) \(10\) c) \(\frac{4}{3}\)

- Complete the operations inside the outer radical first. - Work from the innermost radical outward. - Look for a perfect-square final radicand.

1. For a), \(17^2-15^2=289-225=64\), so \(\sqrt{17^2-15^2}=\sqrt{64}=8\). 2. For b), begin with the inner radical: \(\sqrt{256}=16\). Then \(\sqrt{5\sqrt{256}+1}=\sqrt{5\cdot16+1}=\sqrt{81}=9\).

a) \(8\) b) \(9\)

- Use radical properties to combine or simplify products. - Look for perfect-square factors in larger radicands. - Part c) has the structure of a difference of squares. - What happens when a square root is squared?

1. \(\sqrt{8}\sqrt{2} = \sqrt{16} = 4\), so the result is rational. 2. \(\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}\). Therefore, \(\sqrt{3} + \sqrt{27} = 4\sqrt{3}\), which is irrational. 3. Using the difference of squares, \((5-\sqrt{6})(5+\sqrt{6}) = 5^2 - (\sqrt{6})^2 = 25-6 = 19\), so the result is rational.

a) Rational: \(4\) b) Irrational: \(4\sqrt{3}\) c) Rational: \(19\)

- Combine products or quotients of square roots when possible. - For a difference, evaluate each square root separately.

1. \(\sqrt{3}\cdot\sqrt{27}=\sqrt{81}=9\). 2. \(\frac{\sqrt{72}}{\sqrt{2}}=\sqrt{36}=6\). 3. \(\sqrt{400}-\sqrt{144}=20-12=8\).

a) \(9\) b) \(6\) c) \(8\)

- Combine products or quotients of square roots when their domains permit it. - A radical in a denominator must be defined and nonzero. - Simplify \(\sqrt{x^2}\) using the known sign of \(x\). - A square root multiplied by itself returns the radicand when defined.

1. \(\sqrt{7}\sqrt{28}=\sqrt{196}=14\). No variable restriction is needed. 2. The denominator requires \(x>0\). Then \(\frac{\sqrt{45x^3}}{\sqrt{5x}}=\sqrt{9x^2}=3x\). 3. The radicals require \(a\ge0\). Then \(2\sqrt{a}\cdot3\sqrt{a}=6a\). 4. \(\sqrt{12}+\sqrt{48}=2\sqrt{3}+4\sqrt{3}=6\sqrt{3}\). No variable restriction is needed.

a) \(14\) b) \(3x\), for \(x>0\) c) \(6a\), for \(a\ge0\) d) \(6\sqrt{3}\)

- Square the outside factor when moving it under the radical. - Square both numerator and denominator of a fractional coefficient. - Reverse the usual extraction rule by moving a factor under the radical. - The nonnegative condition ensures the transformation preserves the value.

1. Use \(c\sqrt{d}=\sqrt{c^2d}\) for \(c\ge0\). 2. \(4\sqrt{5}=\sqrt{16\cdot5}=\sqrt{80}\). 3. \(\frac{2}{3}\sqrt{18}=\sqrt{\frac{4}{9}\cdot18}=\sqrt{8}\). 4. \(0.2\sqrt{200}=\sqrt{0.04\cdot200}=\sqrt{8}\). 5. \(k\sqrt{k}=\sqrt{k^2\cdot k}=\sqrt{k^3}\).

a) \(\sqrt{80}\) b) \(\sqrt{8}\) c) \(\sqrt{8}\) d) \(\sqrt{k^3}\)

- Match each expression to a square-of-a-sum or square-of-a-difference formula. - Use \((\sqrt{u})^2=u\) for nonnegative \(u\). - Square every factor in a product. - Combine radical factors in the middle term.

1. \((\sqrt{r}+4)^2=r+8\sqrt{r}+16\). 2. \((5-\sqrt{s})^2=25-10\sqrt{s}+s\). 3. \((\sqrt{3}x+\sqrt{2})^2=3x^2+2\sqrt{6}x+2\). 4. \(\left(\frac{\sqrt{y}}{3}-6\right)^2=\frac{y}{9}-4\sqrt{y}+36\).

a) \(r+8\sqrt{r}+16\) b) \(25-10\sqrt{s}+s\) c) \(3x^2+2\sqrt{6}x+2\) d) \(\frac{y}{9}-4\sqrt{y}+36\)

- Match each trinomial to a squared-binomial pattern. - Identify which terms are the squares of the two binomial terms. - Check whether the middle term is twice the product of the binomial terms. - Remember that \(\sqrt{u^2}=|u|\).

1. In a), \(x^2+6x+9=(x+3)^2\), so the result is \(|x+3|\). 2. In b), \(y^2-10y+25=(y-5)^2\), so the result is \(|y-5|\). 3. In c), \(16+8a+a^2=(a+4)^2\), so the result is \(|a+4|\). 4. In d), \(b^2-2b+1=(b-1)^2\), so the result is \(|b-1|\).

a) \(|x+3|\) b) \(|y-5|\) c) \(|a+4|\) d) \(|b-1|\)

- Can you rewrite the middle term in the same radical form as the others? - Which factors inside the middle square root are perfect squares? - Once the radical terms match, combine their coefficients.

1. Because \(x\ge0\), \(\sqrt{9x^2y}=3x\sqrt{y}\). 2. Combine: \(4x\sqrt{y}-3x\sqrt{y}+2x\sqrt{y}=3x\sqrt{y}\).

\(3x\sqrt{y}\).

- Recognize the product of conjugates. - Square each radical term. - Combine the remaining like terms.

1. Use the difference of squares: \((\sqrt{2p}+\sqrt{3q})(\sqrt{2p}-\sqrt{3q})=2p-3q\). 2. Then \(2p-3q+3q-p=p\).

\(p\)

- A real square root requires a nonnegative radicand. - Reverse the inequality when dividing by a negative number. - Test a value on each side of the boundary.

1. The radicand must be nonnegative: \(24-4x\ge0\). 2. Solving the inequality gives \(-4x\ge-24\), so \(x\le6\).

\(x\le6\)

- Combine radicals with the same index. - Rewrite radicals as rational exponents when the indices differ. - Use the positivity assumptions when simplifying an even root.

1. \(\sqrt3\sqrt{12}=\sqrt{36}=6\). 2. \(\frac{\sqrt{x^3}}{\sqrt{x}}=\sqrt{x^2}=x\), since \(x>0\). 3. \(\sqrt[3]{2}\sqrt[3]{4}=\sqrt[3]{8}=2\). 4. \(\sqrt{a}\sqrt[4]{a}=a^{\frac12}a^{\frac14}=a^{\frac34}=\sqrt[4]{a^3}\).

a) \(6\) b) \(x\) c) \(2\) d) \(\sqrt[4]{a^3}\)

- Combine the fourth roots in the numerator. - Use the quotient property for radicals with the same index. - Rewrite the final radical as a rational exponent.

1. Combine the fourth roots: \(\frac{\sqrt[4]{b^9}\sqrt[4]{b^3}}{\sqrt[4]{b^4}}=\sqrt[4]{\frac{b^{12}}{b^4}}\). 2. Simplify the radicand: \(\sqrt[4]{b^8}=b^2\), since \(b>0\).

\(b^2\)

- Complete operations inside a radical before taking the square root. - Factor out the greatest perfect-square factor. - Apply the quotient property to a square root of a fraction.

1. \(\sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5\). 2. \(\sqrt{72}=\sqrt{36\cdot2}=6\sqrt{2}\). 3. \(\sqrt{\frac{25}{81}}=\frac{\sqrt{25}}{\sqrt{81}}=\frac{5}{9}\).

a) \(5\) b) \(6\sqrt{2}\) c) \(\frac{5}{9}\)

- Combine like terms inside the radical first. - Factor the radicand as a product of perfect squares. - Take the square root of the coefficient and halve the even exponent.

1. Combine like terms inside the radical: \(225z^4-81z^4=144z^4\). 2. Then \(\sqrt{144z^4}=12z^2\).

\(12z^2\)

- Identify radical terms with the same radicand. - A variable directly under a square root must make the radicand nonnegative. - Recall how \(\sqrt{u^2}\) is related to \(u\).

1. In a), combine the coefficients of the like radical terms: \((2+5-1)\sqrt{a}=6\sqrt{a}\). The radicand requires \(a\ge0\). 2. In b), factor out \(x\sqrt{3}\): \((1-0.5)x\sqrt{3}=0.5x\sqrt{3}\). The expression is defined for every real \(x\). 3. In c), \(\sqrt{25z^2}=5|z|\) and \(\sqrt{z^2}=|z|\). Therefore, the expression simplifies to \(6|z|\) and is defined for every real \(z\).

a) \(6\sqrt{a}\), for \(a\ge0\) b) \(0.5x\sqrt{3}\), for \(x\in\mathbb{R}\) c) \(6|z|\), for \(z\in\mathbb{R}\)

- Combine square roots in each product or quotient. - Evaluate the cube root separately. - Add and subtract only after simplifying each part.

1. \(\sqrt2\sqrt{32}=\sqrt{64}=8\). 2. \(\sqrt[3]{27}=3\). 3. \(\frac{\sqrt{50}}{\sqrt2}=\sqrt{25}=5\). 4. Therefore, \(8-3+5=10\).

\(10\)

- For an even-index radical, require the radicand to be nonnegative. - An odd-index radical accepts every real radicand. - Solve each resulting inequality for \(x\).

1. An even root requires \(5x-20\geq0\). Thus \(x\geq4\), so the domain is \([4,\infty)\). 2. A fifth root is defined for every real radicand, so the domain is \((-\infty, \infty)\). 3. An even root requires \(18-3x\geq0\). Thus \(x\leq6\), so the domain is \((-\infty, 6]\). 4. Since \(x^2+9>0\) for every real \(x\), the domain is \((-\infty, \infty)\).

1) \([4,\infty)\) 2) \((-\infty, \infty)\) 3) \((-\infty, 6]\) 4) \((-\infty, \infty)\)

- First rewrite each square root of a square as an absolute value. - Use the given inequality to determine the sign inside the absolute value. - A principal square root is always nonnegative. - Think of absolute value as the nonnegative distance from zero.

1. In a), \(\sqrt{(a-7)^2}=|a-7|\). Since \(a<7\), \(a-7<0\), so \(|a-7|=7-a\). 2. In b), \(\sqrt{(2x-6)^2}=|2x-6|\). Since \(x>3\), \(2x-6>0\), so the result is \(2x-6\). 3. For every \(a<7\), \(7-a>0\). This agrees with the fact that a principal square root is nonnegative.

a) \(7-a\) b) \(2x-6\) In part a, \(7-a>0\) because \(a<7\).

- Factor each radicand into perfect-square factors and any factor left over. - Use the stated conditions to determine the signs of expressions inside absolute values. - Apply the product property of square roots only to nonnegative factors.

1. Rewrite the radicand as \((x-7)^4(x-7)\). Since \(x-7\ge0\), \(\sqrt{(x-7)^4}=(x-7)^2\). Thus, the expression is \((x-7)^2\sqrt{x-7}\). 2. Rewrite the radicand as \(a^2\cdot a\cdot(b+2)^2\). Since \(a\ge0\) and \(b+2\ge0\), \(\sqrt{a^2}=a\) and \(\sqrt{(b+2)^2}=b+2\). Thus, the expression is \(a(b+2)\sqrt{a}\).

1) \((x-7)^2\sqrt{x-7}\) 2) \(a(b+2)\sqrt{a}\)

- An even-index radical requires a nonnegative radicand. - Split a power into a factor whose exponent matches the radical index and a remainder. - Odd-index radicals are defined for every real radicand.

1. The square root requires \(a^5\geq0\), so \(a\geq0\). Then \(\sqrt{a^5}=\sqrt{a^4a}=a^2\sqrt{a}\). 2. A cube root is defined for every real \(x\). Also, \(\sqrt[3]{(x-2)^4}=(x-2)\sqrt[3]{x-2}\). 3. A fifth root is defined for every real \(y\), and \(\sqrt[5]{(y+3)^5}=y+3\).

1) \(a^2\sqrt{a}\), for \(a\geq0\) 2) \((x-2)\sqrt[3]{x-2}\), for \(x\in\mathbb R\) 3) \(y+3\), for \(y\in\mathbb R\)

- Raise an outside coefficient to the radical index before moving it inside. - Once the radical indices match, compare the radicands. - Use the condition \(a\geq0\) in part c.

1. \(3\sqrt7=\sqrt{3^2\cdot7}=\sqrt{63}\), while \(5\sqrt3=\sqrt{5^2\cdot3}=\sqrt{75}\). Since \(75>63\), \(5\sqrt3>3\sqrt7\). 2. \(2\sqrt[3]{5}=\sqrt[3]{2^3\cdot5}=\sqrt[3]{40}\), while \(3\sqrt[3]{2}=\sqrt[3]{3^3\cdot2}=\sqrt[3]{54}\). Since \(54>40\), \(3\sqrt[3]{2}>2\sqrt[3]{5}\). 3. Since \(a\geq0\), \(a^2\sqrt a=\sqrt{(a^2)^2}\sqrt a=\sqrt{a^4a}=\sqrt{a^5}\).

a) \(5\sqrt3>3\sqrt7\) b) \(3\sqrt[3]{2}>2\sqrt[3]{5}\) c) \(a^2\sqrt a=\sqrt{a^5}\)

- Raise each outside factor to the radical index before moving it inside. - Simplify products and quotients of powers after combining them. - Use the positivity assumptions to preserve equivalence.

1. \(4a\sqrt{3a}=\sqrt{(4a)^2\cdot3a}=\sqrt{48a^3}\). 2. \(\frac{x}{y}\sqrt[3]{\frac{y^4}{x^2}}=\sqrt[3]{\left(\frac{x}{y}\right)^3\frac{y^4}{x^2}}=\sqrt[3]{xy}\). 3. \(b^2\sqrt[5]{\frac2{b^7}}=\sqrt[5]{(b^2)^5\frac2{b^7}}=\sqrt[5]{2b^3}\).

a) \(\sqrt{48a^3}\) b) \(\sqrt[3]{xy}\) c) \(\sqrt[5]{2b^3}\)

- Raise each outside factor to the radical index before moving it inside. - Use \((u^p)^q=u^{pq}\). - Add exponents when multiplying powers with the same base.

1. \(a^3\sqrt[5]{a^2}=\sqrt[5]{(a^3)^5a^2}=\sqrt[5]{a^{17}}\). 2. \(xy^2\sqrt[n]{x^2y}=\sqrt[n]{x^ny^{2n}x^2y}=\sqrt[n]{x^{n+2}y^{2n+1}}\). 3. \(b^k\sqrt[3]{b}=\sqrt[3]{(b^k)^3b}=\sqrt[3]{b^{3k+1}}\).

1) \(\sqrt[5]{a^{17}}\) 2) \(\sqrt[n]{x^{n+2}y^{2n+1}}\) 3) \(\sqrt[3]{b^{3k+1}}\)

- Multiply the numerator and denominator so the denominator becomes a perfect power matching the radical index. - Then split the radical of the quotient. - For part d, determine the power of \(3\) needed to make \(3^n\).

1. \(\sqrt{\frac35}=\sqrt{\frac{15}{25}}=\frac{\sqrt{15}}5\). 2. \(\sqrt[3]{\frac29}=\sqrt[3]{\frac6{27}}=\frac{\sqrt[3]{6}}3\). 3. \(\sqrt[4]{\frac18}=\sqrt[4]{\frac2{16}}=\frac{\sqrt[4]{2}}2\). 4. \(\sqrt[n]{\frac13}=\sqrt[n]{\frac{3^{n-1}}{3^n}}=\frac{\sqrt[n]{3^{n-1}}}{3}\).

a) \(\frac{\sqrt{15}}5\) b) \(\frac{\sqrt[3]{6}}3\) c) \(\frac{\sqrt[4]{2}}2\) d) \(\frac{\sqrt[n]{3^{n-1}}}{3}\)

- A positive factor moved under a square root must be squared. - Simplify fractions and variable powers inside the radical. - Check whether the final radicand contains a perfect-square factor.

1. In a), move the positive factor \(6x\) under the radical by squaring it: \(6x\sqrt{\frac{y}{3x}}=\sqrt{(6x)^2\cdot\frac{y}{3x}}=\sqrt{12xy}\). 2. Extract the perfect-square factor \(4\): \(\sqrt{12xy}=2\sqrt{3xy}\). 3. In b), move \(\frac{a}{2}\) under the radical: \(\frac{a}{2}\sqrt{\frac{20b}{a}}=\sqrt{\left(\frac{a}{2}\right)^2\cdot\frac{20b}{a}}=\sqrt{5ab}\).

a) \(2\sqrt{3xy}\) b) \(\sqrt{5ab}\)

- Factor each radicand into a perfect power and a remaining factor. - Match the perfect power to the radical index. - Use the positivity assumptions when extracting variable factors.

1. \(\sqrt{75x^5y^2}=\sqrt{25x^4y^2\cdot3x}=5x^2y\sqrt{3x}\). 2. \(\sqrt[3]{\frac{8a^4}{27b^3}}=\sqrt[3]{\frac{2^3a^3a}{3^3b^3}}=\frac{2a}{3b}\sqrt[3]{a}\).

1) \(5x^2y\sqrt{3x}\) 2) \(\frac{2a}{3b}\sqrt[3]{a}\)

- Factor perfect squares from each radicand. - A like radical must simplify to \(k\sqrt{7}\) for rational \(k\). - Test the quotient of each expression and \(\sqrt{7}\).

1. \(\sqrt{112}=\sqrt{16\cdot7}=4\sqrt{7}\). 2. \(\sqrt{175}=\sqrt{25\cdot7}=5\sqrt{7}\). 3. \(\sqrt{252}=\sqrt{36\cdot7}=6\sqrt{7}\). 4. \(\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4}\). 5. For \(\sqrt{70}\), the quotient \(\frac{\sqrt{70}}{\sqrt{7}}=\sqrt{10}\) is irrational, so \(\sqrt{70}\) is not a rational multiple of \(\sqrt{7}\).

\(\sqrt{70}\) does not belong. The other four expressions are rational multiples of \(\sqrt{7}\).

- Extract the perfect-square factors \(49\) and \(100\). - Simplify inside the parentheses before distributing the outside factor. - Combine the resulting like radical terms.

1. Simplify the radicals: \(\sqrt{49b}=7\sqrt{b}\) and \(\sqrt{100b}=10\sqrt{b}\). 2. Substitute and simplify inside the parentheses: \(3\cdot7\sqrt{b}-2(10\sqrt{b}-4\sqrt{b})=21\sqrt{b}-2(6\sqrt{b})\). 3. Combine like radical terms: \(21\sqrt{b}-12\sqrt{b}=9\sqrt{b}\).

\(9\sqrt{b}\)

- For each term, look for a perfect-cube factor. - What remains inside each cube root after extraction? - Combine coefficients only after the radical parts match.

1. Extract perfect cubes: \(2\sqrt[3]{16}=4\sqrt[3]{2}\), \(\sqrt[3]{54}=3\sqrt[3]{2}\), \(\sqrt[3]{250}=5\sqrt[3]{2}\), and \(\sqrt[3]{128}=4\sqrt[3]{2}\). 2. Combine coefficients: \((4+3-5+4)\sqrt[3]{2}=6\sqrt[3]{2}\).

\(6\sqrt[3]{2}\).

- Extract perfect-square factors from each radicand. - Combine only terms with identical radicands. - Keep track of each sign when removing parentheses.

1. Rewrite each radical: \(2\sqrt{18}=6\sqrt{2}\), \(\sqrt{12}=2\sqrt{3}\), \(\sqrt{8}=2\sqrt{2}\), and \(3\sqrt{27}=9\sqrt{3}\). 2. Combine like radicals: \((6\sqrt{2}+2\sqrt{3})+(2\sqrt{2}-9\sqrt{3})=8\sqrt{2}-7\sqrt{3}\).

\(8\sqrt{2}-7\sqrt{3}\)

- Each product has the form \((u+v)(u-v)\). - When squaring a coefficient times a radical, square both factors. - In c), simplify the product inside the outer radical first.

1. In a), use the difference of squares: \((3\sqrt{5})^2-(2\sqrt{2})^2=45-8=37\). 2. In b), use the same identity: \(\left(\frac{1}{3}\sqrt{a}\right)^2-(\sqrt{b})^2=\frac{a}{9}-b\). 3. In c), first simplify the product under the outer radical: \((\sqrt{13}-2)(\sqrt{13}+2)=13-4=9\). Then \(\sqrt{9}=3\).

a) \(37\) b) \(\frac{a}{9}-b\) c) \(3\)

- Combine radicals with the same index directly. - Rewrite different radical indices as rational exponents. - Add exponents when multiplying and subtract them when dividing.

1. \(\frac{\sqrt[4]{x^5}}{\sqrt[4]{x}}=\sqrt[4]{x^4}=x\). 2. \(\sqrt[3]{a}\sqrt[6]{a}=a^{\frac13+\frac16}=a^{\frac12}=\sqrt a\). 3. \(\frac{\sqrt{y^3}}{\sqrt[4]{y}}=y^{\frac32-\frac14}=y^{\frac54}=y\sqrt[4]{y}\).

1) \(x\) 2) \(\sqrt a\) 3) \(y\sqrt[4]{y}\)

- A square's side length is the square root of its area. - Combine the quotient of square roots under one radical. - A common scale factor in both areas cancels in the ratio.

1. The side-length ratio is \(\frac{a_1}{a_2}=\frac{\sqrt{162}}{\sqrt{2}}=\sqrt{81}=9\), so \(a_1:a_2=9:1\). 2. After doubling both areas, the ratio is \(\frac{\sqrt{2A_1}}{\sqrt{2A_2}}=\sqrt{\frac{A_1}{A_2}}\). The common factor \(2\) cancels, so the ratio remains \(9:1\).

a) \(9:1\) b) The ratio remains \(9:1\).

- Use the quotient property to combine two square roots into one. - Simplify the quotient inside the radical before taking the square root. - In d), apply the quotient rule for powers with the same base. - Use the stated sign condition when simplifying \(\sqrt{a^2}\).

1. In a), \(\frac{\sqrt{180}}{\sqrt{5}}=\sqrt{\frac{180}{5}}=\sqrt{36}=6\). 2. In b), \(\frac{\sqrt{1.2}}{\sqrt{0.3}}=\sqrt{\frac{1.2}{0.3}}=\sqrt{4}=2\). 3. In c), combine the radicals: \(\frac{\sqrt{5\cdot20}}{\sqrt{4}}=\sqrt{\frac{100}{4}}=\sqrt{25}=5\). 4. In d), \(\frac{\sqrt{27a^3}}{\sqrt{3a}}=\sqrt{\frac{27a^3}{3a}}=\sqrt{9a^2}=3a\), because \(a>0\).

a) \(6\) b) \(2\) c) \(5\) d) \(3a\)

- Combine a quotient of square roots under one radical. - Simplify the numerical and variable factors inside the radical. - Divide coefficients outside the radicals separately. - Check whether the final radicand contains a perfect-square factor.

1. \(\frac{\sqrt{108}}{\sqrt{3}}=\sqrt{\frac{108}{3}}=\sqrt{36}=6\). 2. \(\frac{\sqrt{125x^3}}{\sqrt{5x}}=\sqrt{\frac{125x^3}{5x}}=\sqrt{25x^2}=5x\). 3. Divide the coefficients and combine the radicals: \(\frac{12\sqrt{10}}{3\sqrt{2}}=4\sqrt{\frac{10}{2}}=4\sqrt{5}\).

1) \(6\) 2) \(5x\) 3) \(4\sqrt{5}\)

- Rewrite each radicand as a product with a common factor. - Apply the distributive property in reverse. - In part 4), write \(\sqrt{13}=\sqrt{13}\cdot1\).

1. \(\sqrt{15}+\sqrt{10}=\sqrt{5}\sqrt{3}+\sqrt{5}\sqrt{2}=\sqrt{5}(\sqrt{3}+\sqrt{2})\). 2. \(\sqrt{21}-\sqrt{14}=\sqrt{7}(\sqrt{3}-\sqrt{2})\). 3. \(\sqrt{33}+\sqrt{22}=\sqrt{11}(\sqrt{3}+\sqrt{2})\). 4. \(\sqrt{26}-\sqrt{13}=\sqrt{13}(\sqrt{2}-1)\).

1) \(\sqrt{5}(\sqrt{3}+\sqrt{2})\) 2) \(\sqrt{7}(\sqrt{3}-\sqrt{2})\) 3) \(\sqrt{11}(\sqrt{3}+\sqrt{2})\) 4) \(\sqrt{13}(\sqrt{2}-1)\)

- Can you divide each numerator term by the common denominator? - Simplify the quantities inside the resulting square roots. - Use the positivity condition when extracting variable squares.

1. Divide each term in the first numerator by \(\sqrt{5xy}\): \(\sqrt{4x^2}+\sqrt{y^2}=2x+y\), using positive variables. 2. In part 2), \(\frac{\sqrt{a}}{\sqrt{a}}-\frac{\sqrt{4a}}{\sqrt{a}}=1-2=-1\).

1) \(2x+y\). 2) \(-1\).

- Split radicands to reveal a common radical factor. - Rewrite an unrooted variable as a product involving its square root when useful. - Factor just as you would with ordinary algebraic terms.

1. Since \(\sqrt{15a}=\sqrt{5a}\sqrt3\), the expression factors as \(\sqrt{5a}(1+\sqrt3)\). 2. Write \(b=\sqrt b\sqrt b\). Then \(b-\sqrt b=\sqrt b(\sqrt b-1)\). 3. Each term contains \(\sqrt[3]{xy}\): \(\sqrt[3]{x^2y}+\sqrt[3]{xy^2}=\sqrt[3]{xy}(\sqrt[3]{x}+\sqrt[3]{y})\).

1) \(\sqrt{5a}(1+\sqrt3)\) 2) \(\sqrt b(\sqrt b-1)\) 3) \(\sqrt[3]{xy}(\sqrt[3]{x}+\sqrt[3]{y})\)

- Rewrite a nonnegative number or variable as a product of two equal square roots. - Identify the radical factor shared by both terms. - Use the product property to rewrite \(\sqrt{xy}\) when useful. - Use \(\sqrt{u}\sqrt{u}=u\) for \(u\ge0\).

1. Since \(6=\sqrt{6}\cdot\sqrt{6}\), \(6+\sqrt{6}=\sqrt{6}(\sqrt{6}+1)\). 2. Since \(3=\sqrt{3}\cdot\sqrt{3}\), \(3-\sqrt{3}=\sqrt{3}(\sqrt{3}-1)\). 3. Since \(y=\sqrt{y}\cdot\sqrt{y}\), \(y-\sqrt{y}=\sqrt{y}(\sqrt{y}-1)\). 4. Rewrite \(x=\sqrt{x}\cdot\sqrt{x}\) and \(\sqrt{xy}=\sqrt{x}\sqrt{y}\). Then \(x+\sqrt{xy}=\sqrt{x}(\sqrt{x}+\sqrt{y})\).

1) \(\sqrt{6}(\sqrt{6}+1)\) 2) \(\sqrt{3}(\sqrt{3}-1)\) 3) \(\sqrt{y}(\sqrt{y}-1)\) 4) \(\sqrt{x}(\sqrt{x}+\sqrt{y})\)

- Rewrite the nonradical term as the square of a radical. - Identify the common radical factor in the two terms. - Check your factorization by multiplying.

1. Rewrite \(x=(\sqrt{x})^2\), then factor out \(\sqrt{x}\): \(x-\sqrt{x}=\sqrt{x}(\sqrt{x}-1)\). 2. Rewrite \(x+y=(\sqrt{x+y})^2\), then factor out \(\sqrt{x+y}\): \(x+y-\sqrt{x+y}=\sqrt{x+y}(\sqrt{x+y}-1)\).

1) \(\sqrt{x}(\sqrt{x}-1)\) 2) \(\sqrt{x+y}(\sqrt{x+y}-1)\)

- Match each expression to the appropriate binomial-square formula. - Use \((\sqrt{u})^2=u\) for \(u\ge0\). - Square both the coefficient and the radical factor. - Simplify the radical in the middle term.

1. \((3\sqrt{2}+2)^2=18+12\sqrt{2}+4=22+12\sqrt{2}\). 2. \((2-\sqrt{7})^2=4-4\sqrt{7}+7=11-4\sqrt{7}\). 3. \((\sqrt{10}+\sqrt{5})^2=10+2\sqrt{50}+5=15+10\sqrt{2}\).

1) \(22+12\sqrt{2}\) 2) \(11-4\sqrt{7}\) 3) \(15+10\sqrt{2}\)

- Multiply radical indices for nested roots. - Reduce the resulting rational exponent. - Use the positivity assumptions when simplifying variable powers.

1. \(\sqrt[3]{\sqrt{64}}=\sqrt[6]{64}=2\). 2. \(\sqrt[4]{\sqrt{x^2}}=\sqrt[8]{x^2}=x^{\frac14}=\sqrt[4]{x}\). 3. \(\sqrt[3]{\sqrt[5]{y^{10}}}=\sqrt[15]{y^{10}}=y^{\frac23}=\sqrt[3]{y^2}\).

a) \(2\) b) \(\sqrt[4]{x}\) c) \(\sqrt[3]{y^2}\)

- Substitute the given value directly. - Expand the squared binomial. - Look for terms that cancel.

1. Substitute to get \((2+\sqrt{5})^2-4(2+\sqrt{5})+9\). 2. Expand: \((2+\sqrt{5})^2=9+4\sqrt{5}\), and \(-4(2+\sqrt{5})=-8-4\sqrt{5}\). 3. Combining gives \(9+4\sqrt{5}-8-4\sqrt{5}+9=10\).

\(10\)

- A real square root requires a nonnegative radicand. - A square root in a denominator cannot equal zero. - When an expression contains two radicals, intersect their domain conditions. - Combine inequalities and excluded values carefully when writing each domain.

1. For a), require \(5x-15\ge0\), which gives \(x\ge3\). 2. For b), both radicands must be nonnegative. The conditions are \(12-4x\ge0\), so \(x\le3\), and \(x+2\ge0\), so \(x\ge-2\). Their intersection is \(-2\le x\le3\). 3. For c), the radicand in the denominator must be strictly positive: \(2x-8>0\), so \(x>4\).

a) \([3,\infty)\) b) \([-2,3]\) c) \((4,\infty)\)

- Rewrite radicands as powers when useful. - When moving a coefficient into a cube root, how must its power change? - Look for a perfect-cube factor inside the last radical.

1. \(\sqrt[6]{27}=\sqrt[6]{3^3}=3^{\frac{1}{2}}=\sqrt{3}\). 2. Move the coefficient inside the cube root: \(5\sqrt[3]{\frac{2}{25}}=\sqrt[3]{125\cdot\frac{2}{25}}=\sqrt[3]{10}\). 3. Factor out a perfect cube: \(\sqrt[3]{54}=\sqrt[3]{27\cdot2}=3\sqrt[3]{2}\).

a) \(\sqrt{3}\) b) \(\sqrt[3]{10}\) c) \(3\sqrt[3]{2}\)

- Move each positive coefficient under the radical by squaring it. - Compare positive square roots by comparing their radicands. - Rewrite a positive integer as the square root of its square when useful.

1. In a), \(2\sqrt{14}=\sqrt{56}\) and \(3\sqrt{6}=\sqrt{54}\). Since \(56>54\), \(2\sqrt{14}>3\sqrt{6}\). 2. In b), \(0.5\sqrt{80}=\sqrt{20}\) and \(2\sqrt{5}=\sqrt{20}\), so the values are equal. 3. In c), \(4=\sqrt{16}\), and \(16<17\). Therefore, \(4<\sqrt{17}\).

a) \(2\sqrt{14}>3\sqrt{6}\) b) \(0.5\sqrt{80}=2\sqrt{5}\) c) \(4<\sqrt{17}\)

- Determine the sign of the factor outside the radical first. - Rewrite the negative factor as a negative sign times a positive expression. - Square only the positive expression that moves under the radical. - Simplify common factors inside the radicand.

1. Since \(a>1\), \(1-a=-(a-1)\), where \(a-1>0\). 2. Keep the negative sign outside and square the positive factor when moving it under the radical: \(-(a-1)\sqrt{\frac{5}{a-1}}=-\sqrt{(a-1)^2\cdot\frac{5}{a-1}}\). 3. Simplifying the radicand gives \(-\sqrt{5(a-1)}\).

\(-\sqrt{5(a-1)}\)

- Combine the quotient of two square roots into one square root. - Simplify the quotient inside the radical before approximating. - Check the simplified radicand before using a decimal approximation. - Use the hundredths digit to round to the nearest tenth.

1. \(\sqrt{40}\div\sqrt{5}=\sqrt{8}\approx2.8\). 2. \(\sqrt{45}\div\sqrt{3}=\sqrt{15}\approx3.9\). 3. \(\sqrt{21}\div\sqrt{7}=\sqrt{3}\approx1.7\). 4. \(\sqrt{120}\div\sqrt{6}=\sqrt{20}\approx4.5\).

1) \(2.8\) 2) \(3.9\) 3) \(1.7\) 4) \(4.5\)

- Require the expression under the square root to be nonnegative. - For each point, substitute its x-coordinate and compare the result with its y-coordinate. - Check the domain before evaluating a point.

1. The radicand must be nonnegative: \(4x+1\ge0\), so \(x\ge-0.25\). The domain is \([-0.25, \infty)\). 2. For A, \(f(0)=\sqrt{1}-2=-1\), so A is on the graph. 3. For B, \(f(2)=\sqrt{9}-2=1\), so B is on the graph. 4. Point C is not on the graph because \(-1\) is outside the domain. 5. For D, \(f(0.75)=\sqrt{4}-2=0\), so D is on the graph. 6. For E, \(f(6)=\sqrt{25}-2=3\), so E is on the graph.

1. \([-0.25, \infty)\) 2. Points A, B, D, and E lie on the graph. Point C does not.

- Which values are allowed inside a real square root in a denominator? - Substitute each point's x-coordinate into the rule. - Compare the resulting function value with the point's y-coordinate.

1. The denominator \(\sqrt{x}\) is defined and nonzero exactly when \(x>0\); \(g(0)\) is undefined. 2. \(g(1)=4\), \(g(4)=2\), \(g(0.25)=8\), and \(g(16)=1\). Thus, A, B, and D lie on the graph, but C does not.

1) \(x=0\) makes the denominator zero. 2) \(A\), \(B\), and \(D\) only.

- For a negative exponent, rewrite the expression as a reciprocal first. - Interpret the denominator of each exponent as a root index. - Choose the root-before-power order when it makes the arithmetic simpler.

1. \(27^{-\frac{2}{3}}=\frac{1}{27^{\frac{2}{3}}}=\frac{1}{(\sqrt[3]{27})^2}=\frac{1}{9}\). 2. \(16^{\frac{3}{4}}=(\sqrt[4]{16})^3=2^3=8\).

a) \(\frac{1}{9}\) b) \(8\)

- The base is the same in every factor. - Use the product and quotient rules for exponents. - Combine the fractional exponents before simplifying the final fraction.

1. Add exponents in the numerator and subtract the denominator exponent. 2. The combined exponent is \(\frac{5}{6}+\frac{1}{3}-\frac{1}{2}=\frac{5}{6}+\frac{2}{6}-\frac{3}{6}=\frac{4}{6}=\frac{2}{3}\). 3. The simplified expression is \(x^{\frac{2}{3}}\).

\(x^{\frac{2}{3}}\)

- What does the index of the root tell you about the denominator of a fractional exponent? - What role does the power inside the radical play?

1. A fourth root is raising to the power \(\frac{1}{4}\). 2. Thus \(\sqrt[4]{x^3}=(x^3)^{\frac{1}{4}}=x^{\frac{3}{4}}\), so \(p=\frac{3}{4}\).

\(p=\frac{3}{4}\)

- What is required of the number under a real square root? - Is the square of any real number ever negative?

1. For \(f\), the radicand must satisfy \(x-2\ge0\), so \(x\ge2\). 2. The radicand in \(g\) is a square and is nonnegative for every real \(x\), so its domain is all real numbers. 3. Thus \(x=0\) is allowed only in \(g\).

Domain of \(f\): \([2,\infty)\). Domain of \(g\): all real numbers. Only \(g\) allows \(x=0\).

- Are any inputs to the left of the curve's starting point included in its domain? - Trace the requested function output horizontally to the curve, then down to the x-axis.

1. The curve begins at \((1,0)\), so the smallest input in its real domain is \(1\). 2. At height \(2\), the curve has x-coordinate \(5\), giving \(f(5)=2\).

Smallest input: \(1\). The input for \(f(x)=2\) is \(5\).

- How does an outside number change when moved under a square root? - Check by looking for a perfect-square factor in the result.

1. Moving the factor \(3\) under the square root gives \(\sqrt{3^2\cdot2}=\sqrt{18}\). 2. Checking by extraction, \(\sqrt{18}=\sqrt{9\cdot2}=3\sqrt{2}\).

\(\square=18\); \(\sqrt{18}=3\sqrt{2}\).

- Rewrite each radical as a rational exponent. - Add exponents when multiplying powers with the same base. - Separate the whole-number part of the final exponent.

1. Rewrite the radicals as rational exponents: \(a^{\frac43}b^{\frac53}a^{\frac46}b^{\frac36}\). 2. Add exponents with the same base: \(a^{\frac43+\frac23}=a^2\) and \(b^{\frac{10}{6}+\frac36}=b^{\frac{13}{6}}\). 3. Since \(b^{\frac{13}{6}}=b^2b^{\frac16}\), the expression is \(a^2b^2\sqrt[6]{b}\).

d) \(a^2b^2\sqrt[6]{b}\)

- Factor a perfect-square expression from the radicand. - Use the product property of square roots. - The conditions on \(x\) and \(y\) determine how \(\sqrt{x^2}\) simplifies.

1. Factor the greatest perfect-square factor from the radicand: \(8x^2y+4x^3y^2=4x^2(2y+xy^2)\). 2. Then \(\sqrt{4x^2(2y+xy^2)}=\sqrt{4x^2}\sqrt{2y+xy^2}\). 3. Because \(x\ge0\), \(\sqrt{4x^2}=2x\). Therefore, the equivalent expression is \(2x\sqrt{2y+xy^2}\).

a) \(2x\sqrt{2y+xy^2}\)

- Treat each part separately because each uses a different structure. - Positivity lets you simplify roots without sign ambiguity. - Combine matching radicals in parts a and d, and look for a difference of squares in part c.

1. \(\frac{\sqrt{x^3y}}{\sqrt{xy^3}}=\sqrt{\frac{x^3y}{xy^3}}=\sqrt{\frac{x^2}{y^2}}=\frac{x}{y}\), using positivity. 2. \((\sqrt[3]{z^2}\sqrt{z})^6=(z^{\frac23}z^{\frac12})^6=(z^{\frac76})^6=z^7\). 3. \((\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})=x-y\). Then \((x-y)(x+y)=x^2-y^2\). 4. \(\frac{\sqrt[3]{x^2y}}{\sqrt[3]{x^{-1}y^4}}=\sqrt[3]{x^3y^{-3}}=\sqrt[3]{\frac{x^3}{y^3}}=\frac{x}{y}\).

a) \(\frac{x}{y}\) b) \(z^7\) c) \(x^2-y^2\) d) \(\frac{x}{y}\)

- A square's side length is the square root of its area. - Factor a perfect square from each radicand. - Use the rectangle perimeter formula after finding its side lengths. - Like radical terms can be combined as like terms.

1. The side length of square \(A\) is \(a=\sqrt{32}=\sqrt{16\cdot2}=4\sqrt{2}\,\text{cm}\). 2. The side length of square \(B\) is \(b=\sqrt{50}=\sqrt{25\cdot2}=5\sqrt{2}\,\text{cm}\). 3. The rectangle's perimeter is \(P=2(a+b)=2(4\sqrt{2}+5\sqrt{2})=18\sqrt{2}\,\text{cm}\).

a) \(a=4\sqrt{2}\,\text{cm}\) and \(b=5\sqrt{2}\,\text{cm}\) b) \(18\sqrt{2}\,\text{cm}\)

- Convert the mixed numbers to improper fractions. - Decide whether it is easier to simplify under one radical or take separate square roots. - When dividing fractions, multiply by the reciprocal.

1. For a), \(\sqrt{\frac{16}{9}}\cdot\sqrt{\frac{9}{4}}=\frac{4}{3}\cdot\frac{3}{2}=2\). 2. For b), \(\sqrt{\frac{49}{16}}\div\sqrt{\frac{25}{16}}=\frac{7}{4}\div\frac{5}{4}=\frac{7}{5}\). 3. For c), \(\sqrt{\frac{36}{25}\cdot\frac{25}{16}}=\sqrt{\frac{36}{16}}=\frac{3}{2}\).

a) \(2\) b) \(\frac{7}{5}\) c) \(\frac{3}{2}\)

- Evaluate the quantity inside each radical before taking the square root. - Compare the final values, not just the forms of the expressions. - A property that works for multiplication may not work for addition.

1. \(A=\sqrt{144}=12\), and \(B=6\cdot2=12\). Therefore, \(A=B\), illustrating the product property \(\sqrt{ab}=\sqrt{a}\sqrt{b}\) for nonnegative \(a\) and \(b\). 2. \(C=\sqrt{100}=10\), while \(D=8+6=14\). Therefore, \(C\ne D\). 3. A square root does not distribute over addition. In general, \(\sqrt{a+b}\ne\sqrt{a}+\sqrt{b}\).

a) \(A=12\) and \(B=12\); the results are equal. b) \(C=10\) and \(D=14\); the results are not equal. c) In general, \(\sqrt{a+b}\ne\sqrt{a}+\sqrt{b}\).

- Combine like terms inside the radical first. - A square root does not distribute over addition. - Rewrite nested square roots using fractional exponents.

1. For a), \(\sqrt{x^2+x^2}=\sqrt{2x^2}=x\sqrt{2}\) because \(x>0\). Since \(\sqrt{2}\ne2\), the statement is false. 2. For b), \(\sqrt{\sqrt{y^{16}}}=(y^{16})^{1/4}=y^4\).

a) False; the left side equals \(x\sqrt{2}\), not \(2x\). b) \(y^4\)

- A principal square root is never negative. - Rewrite \(\sqrt{u^2}\) using absolute value. - When does \(|x-4|=x-4\)?

1. \(T(1)=\sqrt{(-3)^2}=3\), and \(T(7)=\sqrt{3^2}=3\). 2. For every real expression \(u\), \(\sqrt{u^2}=|u|\). Thus, \(\sqrt{(x-4)^2}=|x-4|\). 3. The equation \(|x-4|=x-4\) is true exactly when \(x-4\ge0\), or \(x\ge4\).

a) \(T(1)=3\) and \(T(7)=3\) b) The equation is true exactly when \(x\ge4\).

- Use \(\sqrt{u^2}=|u|\). - Determine the sign of the expression inside each absolute value. - Compare \(2\) and \(\sqrt{5}\) by comparing their squares.

1. \(\sqrt{(-15)^2}=|-15|=15\). 2. \(\sqrt{(0.4)^2}=|0.4|=0.4\). 3. Since \(2<\sqrt{5}\), \(2-\sqrt{5}<0\). Therefore, \(\sqrt{(2-\sqrt{5})^2}=|2-\sqrt{5}|=\sqrt{5}-2\). 4. For \(y<0\), \(\sqrt{y^2}=|y|=-y\).

a) \(15\) b) \(0.4\) c) \(\sqrt{5}-2\) d) \(-y\)

- Use product and quotient properties of radicals. - Expand the square in part c). - Remember that \(\sqrt{x^2}=|x|\) for real \(x\).

1. \(A=\sqrt{0.04}=0.2=\frac{1}{5}\), so the smallest listed set is \(\mathbb{Q}\). 2. \(B=\sqrt{\frac{27}{3}}=\sqrt{9}=3\), so the smallest listed set is \(\mathbb{N}\). 3. \(C=(1-\sqrt{2})^2=1-2\sqrt{2}+2=3-2\sqrt{2}\), which is irrational. The smallest listed set is \(\mathbb{R}\). 4. \(D=\sqrt{10\cdot0.4}=\sqrt{4}=2\), so the smallest listed set is \(\mathbb{N}\). 5. \(E=\sqrt{(-7)^2}=|-7|=7\), so the smallest listed set is \(\mathbb{N}\).

a) \(0.2\in\mathbb{Q}\) b) \(3\in\mathbb{N}\) c) \(3-2\sqrt{2}\in\mathbb{R}\) d) \(2\in\mathbb{N}\) e) \(7\in\mathbb{N}\)

- Simplify products, differences, or quotients inside a radical before taking the square root. - Use \(\sqrt{u^2}=|u|\) unless the sign of \(u\) is known. - Dividing by a positive decimal less than \(1\) increases the value.

1. In a), \(\sqrt{3a}\sqrt{12a}=\sqrt{36a^2}=6a\), because \(a\ge0\). 2. In b), \(\sqrt{125x^2-44x^2}=\sqrt{81x^2}=9|x|\). 3. In c), \(\sqrt{\frac{16a^2}{0.25}}=\sqrt{64a^2}=8|a|\). 4. In d), \(\sqrt{(-15)^2}=15\) and \(\sqrt{b^2}=|b|\), so the result is \(15+|b|\).

a) \(6a\) b) \(9|x|\) c) \(8|a|\) d) \(15+|b|\)

- Distribute before combining products of square roots. - Simplify square roots of decimal perfect squares. - Use \(\sqrt{u^2}=|u|\). - Cancel a variable factor only when the stated condition makes it nonzero.

1. In a), distribute and combine products of radicals: \(\sqrt{36}+\sqrt{100}=6+10=16\). 2. In b), the numerator is \(0.2|z|\) and the denominator is \(0.1|z|\). Since \(z\ne0\), the quotient is \(2\). 3. In c), \(\sqrt{x^2y^2}=|xy|\), so the difference is \(0\). 4. In d), \(\sqrt{169c^2}=13|c|\) and \(\sqrt{1.69}=1.3\). Therefore, the quotient is \(10|c|\).

a) \(16\) b) \(2\) c) \(0\) d) \(10|c|\)

- A principal square root cannot be negative. - Simplify the quotient inside one radical. - Consider combining the two square roots into a single radical in part c).

1. For \(x=5\), \(\sqrt{5^2}=5\). For \(x=-5\), \(\sqrt{(-5)^2}=5\ne-5\). The correct rule is \(\sqrt{x^2}=|x|\). 2. For \(z>0\), combine the roots: \(\frac{\sqrt{32z^5}}{\sqrt{2z}}=\sqrt{16z^4}=4z^2\). 3. The radical product is \(\sqrt{8}\cdot\sqrt{2}=\sqrt{16}=4\).

a) At \(x=5\), \(\sqrt{5^2}=5\). At \(x=-5\), \(\sqrt{(-5)^2}=5\ne-5\), so the claim is false. The rule is \(\sqrt{x^2}=|x|\). b) \(4z^2\) c) \(4\)

- Simplify powers with the same base inside the radical first. - Taking a square root halves even exponents when the variables are positive. - Use the product property when separating the radical of a product. - Combine the powers of ten before evaluating.

1. Inside the radical, \(\frac{a^4}{a^{-2}}=a^6\), so the expression becomes \(\sqrt{a^6b^6}\). 2. Since \(a\) and \(b\) are positive, \(\sqrt{a^6b^6}=a^3b^3=(ab)^3\). 3. Substituting gives \((10^{-1}\cdot10)^3=(10^0)^3=1\).

Simplified expression: \(a^3b^3\) Value: \(1\)

- Move each positive coefficient under the radical by squaring it. - Compare positive square roots by comparing their radicands. - Rewrite \(0.5\) as \(\frac{1}{2}\) if helpful.

1. In a), \(6\sqrt{2}=\sqrt{36\cdot2}=\sqrt{72}\), while \(5\sqrt{3}=\sqrt{25\cdot3}=\sqrt{75}\). Since \(75>72\), \(5\sqrt{3}>6\sqrt{2}\). 2. In b), \(0.5\sqrt{80}=\sqrt{0.25\cdot80}=\sqrt{20}\), and \(\frac{1}{3}\sqrt{180}=\sqrt{\frac{1}{9}\cdot180}=\sqrt{20}\). The two values are equal.

a) \(5\sqrt{3}>6\sqrt{2}\) b) \(0.5\sqrt{80}=\frac{1}{3}\sqrt{180}\)

- Recognize conjugate products as differences of squares. - Expand the squared sum in part d) before subtracting. - Simplify products of radicals in cross terms.

1. \((\sqrt{11}-\sqrt{d})(\sqrt{11}+\sqrt{d})=11-d\). 2. \((x\sqrt{5}+2y)(x\sqrt{5}-2y)=5x^2-4y^2\). 3. \((\sqrt{8}+\sqrt{2})^2=8+2\sqrt{16}+2=18\). 4. \((\sqrt{a}+\sqrt{b})^2-(a+b)=a+2\sqrt{ab}+b-a-b=2\sqrt{ab}\).

a) \(11-d\) b) \(5x^2-4y^2\) c) \(18\) d) \(2\sqrt{ab}\)

- Compare each expression with \((u\pm v)^2=u^2\pm2uv+v^2\). - Identify \(u\) and \(v\) from the first and middle terms. - The missing term is the square of the second binomial term.

1. In a), let \(u=\sqrt{x}\) and \(v=\sqrt{7}\). The middle term is \(2uv=2\sqrt{7x}\), so the missing term is \(v^2=7\). The expression is \((\sqrt{x}+\sqrt{7})^2\). 2. In b), let \(u=\sqrt{3a}\) and \(v=\sqrt{2}\). Then \(2uv=\sqrt{24a}\), so the missing term is \(v^2=2\). The expression is \((\sqrt{3a}-\sqrt{2})^2\).

a) \(\square=7\), giving \((\sqrt{x}+\sqrt{7})^2\) b) \(\square=2\), giving \((\sqrt{3a}-\sqrt{2})^2\)

- Look for perfect-square trinomials under the radicals. - Do not distribute a square root over a sum. - Rewrite fractional constants as squares when checking a binomial pattern.

1. In a), \(9z^2-12z+4=(3z-2)^2\), so the result is \(|3z-2|\). 2. In b), \(r^2+s^2\) is not a perfect-square trinomial, and a square root does not distribute over addition. It cannot be simplified in general. 3. In c), \(4m^2+20mn+25n^2=(2m+5n)^2\), so the result is \(|2m+5n|\). 4. In d), \(c^2+c+\frac{1}{4}=\left(c+\frac{1}{2}\right)^2\), so the result is \(\left|c+\frac{1}{2}\right|\).

a) \(|3z-2|\) b) Cannot be simplified in general c) \(|2m+5n|\) d) \(\left|c+\frac{1}{2}\right|\)

- Factor out a common numerical factor before looking for a perfect square. - Reorder terms when that makes a binomial-square pattern easier to see. - Treat decimal and fractional coefficients as possible squares.

1. In a), factor \(2\): \(\sqrt{2(x^2+2x+1)}=\sqrt{2(x+1)^2}=\sqrt{2}|x+1|\). 2. In b), \(0.01u^2-0.4uv+4v^2=(0.1u-2v)^2\), so the result is \(|0.1u-2v|\). 3. In c), reorder the terms: \(p^2-4pq+4q^2=(p-2q)^2\), so the result is \(|p-2q|\). 4. In d), the radicand is \(\left(\frac{2}{3}a+b\right)^2\), so the result is \(\left|\frac{2}{3}a+b\right|\).

a) \(\sqrt{2}|x+1|\) b) \(|0.1u-2v|\) c) \(|p-2q|\) d) \(\left|\frac{2}{3}a+b\right|\)

- Substitute the given value into both expressions. - A principal square root cannot be negative. - Square the expression inside the absolute value to build the radicand.

1. At \(x=3\), \(\sqrt{(3-7)^2}=\sqrt{16}=4\), but \(3-7=-4\). The claim is false. The correct identity is \(\sqrt{(x-7)^2}=|x-7|\). 2. To obtain \(|4a-3b|\), the radicand must be \((4a-3b)^2\). 3. Expanding gives \((4a-3b)^2=16a^2-24ab+9b^2\).

a) The claim is false because \(4\ne-4\). b) \(16a^2-24ab+9b^2\)

- Match the radicand to a squared-binomial pattern. - A square of a real expression is never negative. - Opposite numbers have the same absolute value.

1. The radicand is a perfect square: \(\frac{1}{9}x^2-\frac{2}{3}x+1=\left(\frac{x}{3}-1\right)^2\). Thus, \(T(x)=\left|\frac{x}{3}-1\right|\). 2. A squared real expression is always nonnegative, so the radicand is defined for every real \(x\). The domain is \(\mathbb{R}\). 3. \(\sqrt{(x-3)^2}=|x-3|\) and \(\sqrt{(3-x)^2}=|3-x|\). Since \(3-x=-(x-3)\), the absolute values are equal for every real \(x\).

a) \(\left|\frac{x}{3}-1\right|\) b) \(\mathbb{R}\) c) Yes. Both expressions equal \(|x-3|\).

- How can you separate a perfect-square factor from the first radicand? - Check whether the numerator contains like radical terms. - Why is the denominator nonzero under the stated condition?

1. Because \(a>0\), \(\sqrt{a^3}=a\sqrt{a}\). 2. The numerator is \(3a\sqrt{a}\), so the quotient is \(3a\). 3. Subtract \(a\) to get \(2a\).

\(2a\).

- Apply both the radical restriction and the denominator restriction. - Determine when a fraction with positive numerator is positive. - The denominator cannot equal zero.

1. The fraction under the radical must be nonnegative, and its denominator cannot be zero. 2. Because the numerator \(5\) is positive, the quotient is nonnegative only when \(a-3>0\). 3. Therefore, \(a>3\).

\(a>3\), or \((3,\infty)\)

- Squares are always nonnegative. - A real square root requires a nonnegative radicand. - Check separately whether any denominator can equal zero.

1. For a), \(k^2+10>0\) for every real \(k\), so the expression is defined for all real numbers. 2. For b), \(-(k-2)^2\le0\). It is nonnegative only when \((k-2)^2=0\), so \(k=2\). 3. For c), \(k\ne0\) is required by the denominator. For every nonzero \(k\), \(\frac{4}{k^2}>0\), so the expression is defined.

a) \(k\in\mathbb{R}\) b) \(k=2\) c) \(k\ne0\)

- Rewrite constants as perfect powers matching the radical index. - Multiply exponents when raising a power to another power. - Combine radicals with the same index before simplifying.

1. Since \(81=3^4\), \(\sqrt[4]{81a^{12}}=\sqrt[4]{3^4a^{12}}=3a^3\). 2. \(\sqrt{\sqrt[3]{x^{18}}}=x^{18\cdot\frac13\cdot\frac12}=x^3\). 3. \(\frac{\sqrt[5]{b^{13}}}{\sqrt[5]{b^3}}=\sqrt[5]{b^{10}}=b^2\). 4. \((\sqrt[3]{y})^{-6}=(y^{\frac13})^{-6}=y^{-2}=\frac1{y^2}\).

a) \(3a^3\) b) \(x^3\) c) \(b^2\) d) \(y^{-2}=\frac1{y^2}\)

- Rewrite each radical as a power with exponent \(\frac1n\). - Apply the product property for powers with the same exponent. - In part b, combine the radicands first.

1. Rewrite the radicals: \(\sqrt[n]{a}\sqrt[n]{b}=a^{\frac1n}b^{\frac1n}\). 2. Apply the property \(u^rv^r=(uv)^r\): \(a^{\frac1n}b^{\frac1n}=(ab)^{\frac1n}=\sqrt[n]{ab}\). 3. \(\sqrt[4]{2}\sqrt[4]{8}=\sqrt[4]{16}=2\).

a) \(a^{\frac1n}b^{\frac1n}=(ab)^{\frac1n}=\sqrt[n]{ab}\) b) \(2\)

- Combine matching radical indices directly. - For different indices, rewrite the radicals as rational exponents. - Add exponents when multiplying powers with the same base and subtract them when dividing.

1. \(\sqrt[n]{x^2}\sqrt[n]{x^3}=\sqrt[n]{x^5}\). 2. \(\frac{\sqrt[3]{y^2}}{\sqrt[6]{y}}=y^{\frac23-\frac16}=y^{\frac12}=\sqrt{y}\). 3. \(\sqrt{x}\sqrt[3]{x}\sqrt[6]{x}=x^{\frac12+\frac13+\frac16}=x^1=x\).

a) \(\sqrt[n]{x^5}\) b) \(\sqrt{y}\) c) \(x\)

- Combine the quotient under one cube root. - Use the simplified expression in part b. - In part c, compare the rational exponents before and after the proposed change.

1. \(T=\sqrt[3]{\frac{z^7}{z}}=\sqrt[3]{z^6}=z^2\). 2. For \(z=8\), \(T=8^2=64\). 3. The original expression is \(\sqrt[4]{x^{12}}=x^3\). Doubling the index and halving the exponent gives \(\sqrt[8]{x^6}=x^{\frac{6}{8}}=x^{\frac{3}{4}}\). These are not equal for all positive \(x\), so the claim is false.

a) \(z^2\) b) \(64\) c) The claim is false: \(\sqrt[4]{x^{12}}=x^3\), but \(\sqrt[8]{x^6}=x^{\frac{3}{4}}\).

- Combine the product under one radical. - Write \(1024\) as a power of \(2\). - Determine when the resulting exponent is a positive integer.

1. Combine the radicals: \(\sqrt[n]{128}\sqrt[n]{8}=\sqrt[n]{1024}\). 2. Since \(1024=2^{10}\), the expression is \(2^{\frac{10}{n}}\). 3. For this value to be an integer, \(\frac{10}{n}\) must be a positive integer. Thus \(n\) must be a divisor of \(10\). 4. The divisors of \(10\) greater than \(1\) are \(2\), \(5\), and \(10\).

\(n\in\{2,5,10\}\)

- A root of a root multiplies the rational exponents. - Simplify the numerator and denominator of \(T_2\) under one square root. - Compare the two final powers of \(a\).

1. \(T_1=(a^{18})^{\frac12\cdot\frac13}=a^3\). 2. \(T_2=\sqrt{\frac{a^5a^7}{a^6}}=\sqrt{a^6}=a^3\), since \(a>0\). 3. Both expressions simplify to \(a^3\), so they are equal for every positive \(a\).

Yes. Both expressions simplify to \(a^3\).

- Simplify from the innermost radical outward. - Use the assumption \(x>0\) when taking even roots. - Combine powers with the same base before simplifying each radical.

1. Start with the innermost radical: \(\sqrt{x^4}=x^2\). 2. Then \(\sqrt[4]{x^2\cdot x^2}=\sqrt[4]{x^4}=x\). 3. Finally, \(\sqrt{x^3\cdot x}=\sqrt{x^4}=x^2\).

\(x^2\)

- Extract perfect-square factors from each radicand. - Use the quotient property for radicals when the quantities are defined. - Apply \(\sqrt{u^2}=|u|\). - Opposite expressions have equal absolute values.

1. In a), \(\sqrt{18x^2}=\sqrt{9\cdot2\cdot x^2}=3\sqrt{2}|x|\). 2. In b), combine the quotient under one radical: \(\sqrt{\frac{12y^3}{3y}}=\sqrt{4y^2}=2|y|=2y\), because \(y>0\). 3. In c), the expression becomes \(|a-b|-|b-a|\). Since \(b-a=-(a-b)\), the two absolute values are equal, so the result is \(0\).

a) \(3\sqrt{2}|x|\) b) \(2y\) c) \(0\)

- Simplify each radical before combining terms. - Combine like radical terms inside each set of parentheses. - Follow the order of operations.

1. Since \(\sqrt{8}=2\sqrt{2}\), the first term is \((3\sqrt{2}-2\sqrt{2})^2=(\sqrt{2})^2=2\). 2. Since \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), the second term is \(\sqrt{3}(2\sqrt{3}+3\sqrt{3})=\sqrt{3}(5\sqrt{3})=15\). 3. Adding gives \(2+15=17\).

\(17\)

- Rewrite \(\sqrt{12}\) in simplest radical form. - Use radical properties for multiplication and division. - Use a squared-binomial identity in the expression that contains one. - Simplify before classifying each result.

1. \(a+b=\sqrt{3}+\sqrt{12}=\sqrt{3}+2\sqrt{3}=3\sqrt{3}\), which is irrational. 2. \(ab=\sqrt{3}\sqrt{12}=\sqrt{36}=6\), which is rational. 3. \(b\div a=\sqrt{12}\div\sqrt{3}=\sqrt{4}=2\), which is rational. 4. \(a^2+b^2=3+12=15\), which is rational. 5. \((b-a)^2=(2\sqrt{3}-\sqrt{3})^2=(\sqrt{3})^2=3\), which is rational.

1) \(3\sqrt{3}\), irrational 2) \(6\), rational 3) \(2\), rational 4) \(15\), rational 5) \(3\), rational

- Evaluate each side of line 5 separately. - Recall that a principal square root is nonnegative. - Apply \(\sqrt{u^2}=|u|\), especially when \(u\) may be negative. - Check how the rule behaves when the quantity being squared is negative.

1. Lines 1 through 4 are valid because both squared expressions equal \(1\). 2. The error occurs in line 5. In general, \(\sqrt{u^2}=|u|\), not \(u\). 3. The correct line is \(|2-3|=|4-3|\), which simplifies to \(1=1\). It does not imply \(2=4\).

The error is in line 5. It should be \(|2-3|=|4-3|\), because \(\sqrt{u^2}=|u|\).

- Rewrite the radicand using a squared-binomial pattern. - Substitute the given values into both the original expression and the proposed simplification. - A principal square root cannot be negative. - Determine when \(x-5\) is nonnegative.

1. The radicand is a perfect square: \(x^2-10x+25=(x-5)^2\). 2. For \(x=8\), \(T(8)=\sqrt{(8-5)^2}=\sqrt{3^2}=3\). 3. For \(x=2\), \(T(2)=\sqrt{(2-5)^2}=\sqrt{(-3)^2}=3\). 4. The proposed expression gives \(8-5=3\), so it works when \(x=8\). It gives \(2-5=-3\), so it does not work when \(x=2\). 5. In general, \(T(x)=\sqrt{(x-5)^2}=|x-5|\). This equals \(x-5\) exactly when \(x-5\ge0\), or \(x\ge5\).

a) \(T(8)=3\) and \(T(2)=3\) b) The claim is true for \(x=8\) but false for \(x=2\). The simplification \(T(x)=x-5\) is correct for \(x\ge5\).

- Rewrite each radicand as the square of a binomial. - Use the relationship between the square root of a square and absolute value. - Determine the signs of \(x+2\) and \(x-2\) on the given interval. - Use those signs to remove the absolute-value symbols.

1. Rewrite each radicand as a perfect square: \(T(x)=\sqrt{(x+2)^2}-\sqrt{(x-2)^2}\). 2. Apply \(\sqrt{u^2}=|u|\): \(T(x)=|x+2|-|x-2|\). 3. Because \(-2<x<2\), \(x+2>0\) and \(x-2<0\). Therefore, \(|x+2|=x+2\) and \(|x-2|=2-x\). 4. Thus, \(T(x)=(x+2)-(2-x)=2x\).

\(2x\)

- Rewrite the radicand as the square of a binomial. - Recall how absolute value changes when its input is negative. - Treat the two given ranges of \(y\) separately.

1. Rewrite the radicand as a perfect square: \(y^2-10y+25=(y-5)^2\). Therefore, \(E(y)=\frac{y+|y-5|}{2}\). 2. When \(y\ge5\), \(|y-5|=y-5\). Then \(E(y)=\frac{y+y-5}{2}=\frac{2y-5}{2}=y-2.5\). 3. When \(y<5\), \(|y-5|=5-y\). Then \(E(y)=\frac{y+5-y}{2}=\frac{5}{2}=2.5\).

1) For \(y\ge5\): \(y-2.5\) 2) For \(y<5\): \(2.5\)

- Replace the square root of each squared expression with an absolute value. - Use the stated sign conditions to remove the absolute-value symbols. - In part c, determine when a number’s absolute value equals its opposite.

1. In a), \(\sqrt{x^2\cdot y^4}=\sqrt{x^2}\sqrt{(y^2)^2}=|x||y^2|\). Since \(y^2\ge0\), \(|y^2|=y^2\). Since \(x<0\), \(|x|=-x\). The result is \(-xy^2\). 2. In b), \(\sqrt{\frac{25a^2}{b^2}}=\frac{5|a|}{|b|}\). Because \(a>0\), \(|a|=a\), and because \(b<0\), \(|b|=-b\). Therefore, the result is \(\frac{5a}{-b}=-\frac{5a}{b}\). 3. In c), \(\sqrt{z^2}=|z|\), so the equation is \(|z|=-z\). This is true exactly when \(z\le0\).

a) \(-xy^2\) b) \(-\frac{5a}{b}\) c) \(z\le0\)

- Recall what happens when you take the principal square root of a squared expression. - A principal square root is never negative. - Compare \(5\) with \(\sqrt{26}\) without using a calculator. - Extract the perfect-square factor from \(\sqrt{12}\).

1. In a), \(\sqrt{(5-\sqrt{26})^2}=|5-\sqrt{26}|\). 2. Since \(5=\sqrt{25}<\sqrt{26}\), the expression \(5-\sqrt{26}\) is negative. Therefore, \(|5-\sqrt{26}|=\sqrt{26}-5\). 3. In b), \(\sqrt{12(\sqrt{3}-2)^2}=\sqrt{12}|\sqrt{3}-2|\). 4. Since \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{3}<2\), \(|\sqrt{3}-2|=2-\sqrt{3}\). 5. Thus, \(2\sqrt{3}(2-\sqrt{3})=4\sqrt{3}-6\).

a) \(\sqrt{26}-5\) b) \(4\sqrt{3}-6\)

- Separate the largest even power of \(a\) from the factor left under the radical. - A square root of a squared expression produces an absolute value. - Use the condition \(b<3\) to simplify that absolute value.

1. Rewrite the radicand as \(a^6\cdot a\cdot(b-3)^2\). 2. Since \(a\ge0\), \(\sqrt{a^6}=a^3\). 3. Because \(b<3\), \(b-3<0\), so \(\sqrt{(b-3)^2}=|b-3|=3-b\). 4. Therefore, the simplified expression is \(a^3(3-b)\sqrt{a}\).

\(a^3(3-b)\sqrt{a}\)

- Identify factors whose exponents are multiples of \(3\). - Recall that cube roots of negative numbers are real. - For an odd index, \(\sqrt[n]{u^n}=u\) without an absolute value.

1. Rewrite the radicand as \(\frac{(x^2)^3y^3y}{3^3z^3}\). 2. Extract the perfect cubes: \(A=\frac{x^2y\sqrt[3]{y}}{3z}\). 3. A cube root is defined for every real radicand, and \(\sqrt[3]{u^3}=u\) for every real \(u\). Therefore the simplification remains valid for negative \(y\) or \(z\), provided \(z\neq0\).

a) \(\frac{x^2y\sqrt[3]{y}}{3z}\) b) Yes. It is valid for all real \(x\) and \(y\) and all real \(z\neq0\).

- Factor each radicand into a perfect power and a remaining factor. - Match the perfect-power exponent to the radical index. - For an even-index radical, use the domain to determine whether an absolute value is needed.

1. \(\sqrt[3]{54a^4}=\sqrt[3]{27a^3\cdot2a}=3a\sqrt[3]{2a}\). 2. \(\sqrt[3]{16(x+2)^5}=\sqrt[3]{8(x+2)^3\cdot2(x+2)^2}=2(x+2)\sqrt[3]{2(x+2)^2}\). 3. For the fourth root to be real, \(y\geq0\). Then \(\sqrt[4]{80y^7}=\sqrt[4]{16y^4\cdot5y^3}=2y\sqrt[4]{5y^3}\).

1) \(3a\sqrt[3]{2a}\) 2) \(2(x+2)\sqrt[3]{2(x+2)^2}\) 3) \(2y\sqrt[4]{5y^3}\)

- Combine fractions before simplifying a radical. - Factor out common powers whose exponents are multiples of the radical index. - Use the stated domain conditions when extracting even powers.

1. \(\sqrt[3]{\frac{8b^4}{27c^6}}=\sqrt[3]{\frac{2^3b^3b}{3^3(c^2)^3}}=\frac{2b}{3c^2}\sqrt[3]{b}\). 2. Combine the fractions: \(\frac1{a^8}+\frac1{a^9}=\frac{a+1}{a^9}=\frac1{a^8}\cdot\frac{a+1}{a}\). Since \(a^2>0\), the fourth root is \(\frac1{a^2}\sqrt[4]{\frac{a+1}{a}}\). 3. \(\sqrt[3]{a^6(b-c)}=a^2\sqrt[3]{b-c}\).

1) \(\frac{2b}{3c^2}\sqrt[3]{b}\) 2) \(\frac1{a^2}\sqrt[4]{\frac{a+1}{a}}\) 3) \(a^2\sqrt[3]{b-c}\)

- Factor each radicand before extracting powers. - Odd-index radicals allow negative radicands. - For part 3, factor out the greatest common power of \(x\) and analyze the sign of the remaining factor.

1. The fourth root requires \(b\geq2\). Then \(\sqrt[4]{48(b-2)^5}=\sqrt[4]{16(b-2)^4\cdot3(b-2)}=2(b-2)\sqrt[4]{3(b-2)}\). 2. The cube root is defined for every real \(c\). Factoring gives \(\sqrt[3]{-8(c+1)^6\cdot3(c+1)}=-2(c+1)^2\sqrt[3]{3(c+1)}\). 3. Factor the radicand: \(x^5+x^4=x^4(x+1)\). It is nonnegative exactly when \(x\geq-1\). Thus \(\sqrt{x^4(x+1)}=x^2\sqrt{x+1}\).

1) \(2(b-2)\sqrt[4]{3(b-2)}\), for \(b\geq2\) 2) \(-2(c+1)^2\sqrt[3]{3(c+1)}\), for \(c\in\mathbb R\) 3) \(x^2\sqrt{x+1}\), for \(x\geq-1\)

- A positive coefficient moved under a square root must be squared. - Compare positive radicals by comparing their radicands. - Remember that a principal square root is never negative. - Use \(\sqrt{x^2}=|x|\).

1. In a), \(4\sqrt{3}=\sqrt{48}\) and \(3\sqrt{5}=\sqrt{45}\), so \(4\sqrt{3}>3\sqrt{5}\). 2. In b), \(2\sqrt{10}=\sqrt{40}\) and \(3\sqrt{4.5}=\sqrt{40.5}\), so \(2\sqrt{10}<3\sqrt{4.5}\). 3. For \(x=-3\), the left side is \(-3\sqrt{2}\), while the right side is \(\sqrt{2(-3)^2}=\sqrt{18}=3\sqrt{2}\). They are not equal. 4. In general, \(\sqrt{2x^2}=|x|\sqrt{2}\), not \(x\sqrt{2}\). The two expressions agree only when \(x\ge0\).

1. a) \(4\sqrt{3}>3\sqrt{5}\) b) \(2\sqrt{10}<3\sqrt{4.5}\) 2. The claim is false. At \(x=-3\), \(-3\sqrt{2}\ne3\sqrt{2}\). In general, \(\sqrt{2x^2}=|x|\sqrt{2}\).

- A positive factor moved under a square root must be squared. - Simplify numerical factors and variable powers after combining them under the radical. - Use \((mn)^2=m^2n^2\) when moving a product under a radical. - Use the assumption that every variable is positive.

1. In a), move \(\frac{a}{2}\) under the radical by squaring it: \(\sqrt{\left(\frac{a}{2}\right)^2\cdot\frac{8}{a}}=\sqrt{2a}\). 2. In b), move \(3b^2\) under the radical: \(\sqrt{(3b^2)^2\cdot\frac{1}{3b^3}}=\sqrt{3b}\). 3. In c), move \(\frac{2}{3x}\) under the radical: \(\sqrt{\left(\frac{2}{3x}\right)^2\cdot18x^3}=\sqrt{8x}\).

a) \(\sqrt{2a}\) b) \(\sqrt{3b}\) c) \(\sqrt{8x}\)

- Compare an outside factor with the radical index before moving it inside. - Check how both numerical coefficients and variable powers are affected by that move. - For an even-index radical, keep the sign convention for the principal root in mind.

1. \(x\sqrt[4]{x^3}=\sqrt[4]{x^4x^3}=\sqrt[4]{x^7}\). 2. The outside factor must be cubed: \(2y^2\sqrt[3]{3y}=\sqrt[3]{(2y^2)^3\cdot3y}=\sqrt[3]{24y^7}\). The claim is false. 3. If \(a<0\) and \(b>0\), then \(a\sqrt{b}<0\), but \(\sqrt{a^2b}\ge0\). Moving a negative factor inside an even-index radical loses its sign.

a) \(\sqrt[4]{x^7}\) b) The claim is false; the correct expression is \(\sqrt[3]{24y^7}\). c) For \(a<0\) and \(b>0\), the left side is negative while the right side is nonnegative.

- Cube the entire outside factor in part a. - Check whether a transformation preserves the sign of the original expression. - A principal square root cannot be negative.

1. \(\frac{k}{2}\sqrt[3]{\frac{24}{k^2}}=\sqrt[3]{\left(\frac{k}{2}\right)^3\frac{24}{k^2}}=\sqrt[3]{3k}\). 2. The original expression is negative, but \(\sqrt{18}\) is nonnegative. A negative sign cannot be moved inside a square root by squaring the coefficient. The correct transformation is \(-3\sqrt{2}=-\sqrt{18}\).

a) \(\sqrt[3]{3k}\) b) The claim is false. The correct expression is \(-\sqrt{18}\).

- Split each exponent into a multiple of the radical index plus a remainder. - Extract factors whose exponents are divisible by the radical index. - Simplify matching powers in a numerator and denominator.

1. \(\sqrt[4]{x^9y^4}=\sqrt[4]{x^8y^4\cdot x}=x^2y\sqrt[4]{x}\). 2. \(\sqrt[n]{a^{n+2}b^{3n}}=\sqrt[n]{a^na^2(b^3)^n}=ab^3\sqrt[n]{a^2}\). 3. \(\frac{\sqrt[k]{z^{2k+1}}}{z}=\frac{z^2\sqrt[k]{z}}{z}=z\sqrt[k]{z}\).

1) \(x^2y\sqrt[4]{x}\) 2) \(ab^3\sqrt[n]{a^2}\) 3) \(z\sqrt[k]{z}\)

- Raise the outside factor to the radical index before moving it inside; for the square root, first note that \(2x^2>0\) on the stated domain. - In part b, recognize a perfect-square trinomial. - Simplify products and quotients inside the radical after combining them.

1. \(2x^2\sqrt{\frac1{4x^3}+\frac3{2x^4}}=\sqrt{4x^4\left(\frac1{4x^3}+\frac3{2x^4}\right)}=\sqrt{x+6}\). 2. Since \(a^2-2ab+b^2=(a-b)^2\), \(\frac{a-b}{a}\sqrt[3]{\frac{a^4}{(a-b)^2}}=\sqrt[3]{\left(\frac{a-b}{a}\right)^3\frac{a^4}{(a-b)^2}}=\sqrt[3]{a(a-b)}\).

a) \(\sqrt{x+6}\) b) \(\sqrt[3]{a(a-b)}\)

- Does the polynomial in part a) match a familiar binomial power? - Is there a common factor in both terms of part b)'s radicand? - What power must an outside factor have when moved inside a cube root?

1. The polynomial in a) is \((k+1)^3\). Since \(k+1>0\), \(\frac{1}{k+1}\sqrt{(k+1)^3}=\sqrt{k+1}\). 2. Factor the cube-root radicand in b) as \(\frac{y^3(y+1)}{x^2}\). 3. Move \(\frac{x}{y}\) into the cube root as \(\frac{x^3}{y^3}\). The result is \(\sqrt[3]{x(y+1)}\).

a) \(\sqrt{k+1}\). b) \(\sqrt[3]{x(y+1)}\).

- Factor denominators using identities such as the difference of squares. - Raise an outside fraction to the radical index before moving it inside. - Look for a perfect-square trinomial in part c.

1. Since \(x^2-y^2=(x+y)(x-y)\), \((x+y)\sqrt{\frac1{x^2-y^2}}=\sqrt{\frac{(x+y)^2}{(x+y)(x-y)}}=\sqrt{\frac{x+y}{x-y}}\). 2. \(\frac{a^2}{b}\sqrt[3]{\frac{b^4}{a^5}}=\sqrt[3]{\frac{a^6}{b^3}\cdot\frac{b^4}{a^5}}=\sqrt[3]{ab}\). 3. Since \(m^2-10m+25=(m-5)^2\), \((m-5)\sqrt[4]{\frac1{(m-5)^2}}=\sqrt[4]{\frac{(m-5)^4}{(m-5)^2}}=\sqrt[4]{(m-5)^2}=\sqrt{m-5}\).

a) \(\sqrt{\frac{x+y}{x-y}}\) b) \(\sqrt[3]{ab}\) c) \(\sqrt{m-5}\)

- Raise the outside factor to the radical index. - Multiply powers and then subtract exponents when canceling. - Use the positivity assumptions to move factors inside an even-index radical.

1. \(x\sqrt{\frac3x}=\sqrt{x^2\cdot\frac3x}=\sqrt{3x}\). 2. \(a\sqrt[3]{\frac{b}{a^2}}=\sqrt[3]{a^3\cdot\frac{b}{a^2}}=\sqrt[3]{ab}\). 3. \(y^2\sqrt[4]{\frac1{y^7}}=\sqrt[4]{y^8\cdot\frac1{y^7}}=\sqrt[4]{y}\).

a) \(\sqrt{3x}\) b) \(\sqrt[3]{ab}\) c) \(\sqrt[4]{y}\)

- First move the outside variable inside the radical. - Then make the denominator a perfect power matching the radical index. - Extract the radical of that denominator.

1. Move \(m\) inside the cube root: \(m\sqrt[3]{\frac{n}{2m^2}}=\sqrt[3]{\frac{mn}{2}}\). Multiply the fraction by \(\frac4{4}\): \(\sqrt[3]{\frac{4mn}{8}}=\frac12\sqrt[3]{4mn}\). 2. Move \(y\) inside the fourth root: \(y\sqrt[4]{\frac{x}{8y^3}}=\sqrt[4]{\frac{xy}{8}}\). Multiply the fraction by \(\frac2{2}\): \(\sqrt[4]{\frac{2xy}{16}}=\frac12\sqrt[4]{2xy}\).

a) \(\frac12\sqrt[3]{4mn}\) b) \(\frac12\sqrt[4]{2xy}\)

- Move the outside factor inside by raising it to the radical index. - Simplify the resulting powers and numerical factors. - Extract any remaining perfect square or perfect cube.

1. \(b^2\sqrt{\frac{18}{b^3}}=\sqrt{b^4\cdot\frac{18}{b^3}}=\sqrt{18b}=3\sqrt{2b}\). 2. \(\frac{x}{3}\sqrt[3]{\frac{54}{x^2}}=\sqrt[3]{\frac{x^3}{27}\cdot\frac{54}{x^2}}=\sqrt[3]{2x}\).

a) \(3\sqrt{2b}\) b) \(\sqrt[3]{2x}\)

- Factor out the greatest common numerical factor and the smallest power of \(x\). - Look for the pattern \(a^2-2ab+b^2\) in the remaining trinomial. - Use \(\sqrt{u^2}=|u|\) when extracting a squared factor. - Determine when the entire radicand is nonnegative.

1. Factor \(2x\) from the radicand: \(18x^3-24x^2+8x=2x(9x^2-12x+4)\). 2. The trinomial is a perfect square: \(9x^2-12x+4=(3x-2)^2\). 3. Therefore, \(T(x)=\sqrt{2x(3x-2)^2}=|3x-2|\sqrt{2x}\). 4. For the original radical to be real, \(2x(3x-2)^2\ge0\). The squared factor is nonnegative, and its zero occurs at \(x=\frac{2}{3}\), which is already nonnegative. For every \(x<0\), the product is negative. Thus, the domain is \(x\ge0\).

a) \(2x(3x-2)^2\) b) \(T(x)=|3x-2|\sqrt{2x}\); domain: \([0,\infty)\)

- Cube the outside factor before moving it inside. - Distribute that factor across every term in the radicand. - Simplify powers in each resulting fraction.

1. \(2x\sqrt[3]{1+\frac1{8x^3}}=\sqrt[3]{8x^3\left(1+\frac1{8x^3}\right)}=\sqrt[3]{8x^3+1}\). 2. \(\frac{a}{b}\sqrt[3]{\frac{b^3}{a^2}-\frac{b^4}{a^3}}=\sqrt[3]{\frac{a^3}{b^3}\left(\frac{b^3}{a^2}-\frac{b^4}{a^3}\right)}=\sqrt[3]{a-b}\).

1) \(\sqrt[3]{8x^3+1}\) 2) \(\sqrt[3]{a-b}\)

- Move the outside fraction into the cube root. - Cube both its numerator and denominator. - Distribute and simplify the two terms in the radicand.

1. Move \(\frac{x}{3}\) inside the cube root by cubing it: \(A=\sqrt[3]{\frac{x^3}{27}\left(\frac{54}{x^3}+\frac{27}{x^2}\right)}\). 2. Simplify the radicand: \(\frac{x^3}{27}\cdot\frac{54}{x^3}+\frac{x^3}{27}\cdot\frac{27}{x^2}=2+x\). 3. Thus \(A=\sqrt[3]{x+2}=B\), so the claim is true.

The claim is true: \(A=\sqrt[3]{x+2}=B\).

- Rewrite the outside factor as an \(n\)th root. - Combine the two radicals. - Subtract exponents when dividing powers with the same base.

1. Rewrite \(a^2\) as an \(n\)th root: \(a^2=\sqrt[n]{a^{2n}}\). 2. Combine the radicals and simplify the power of \(a\): \(\sqrt[n]{a^{2n}\cdot\frac{b}{a^{2n-1}}}=\sqrt[n]{ab}\).

\(\sqrt[n]{ab}\)

- Make the denominator exponent equal to \(n\). - Multiply both numerator and denominator by the same factor. - Then extract the perfect \(n\)th power from the denominator.

1. Multiply the radicand fraction by \(\frac{(x+y)^2}{(x+y)^2}\) so the denominator becomes an exact \(n\)th power: \(\sqrt[n]{\frac{xy^2(x+y)^2}{(x+y)^n}}\). 2. Since \(x+y>0\), extract the denominator: \(\frac1{x+y}\sqrt[n]{xy^2(x+y)^2}\).

\(\frac1{x+y}\sqrt[n]{xy^2(x+y)^2}\)

- Consider what happens when a negative number is squared and then its principal square root is taken. - A principal square root is always nonnegative. - Rewrite the radical using the definition of absolute value. - Test one value on each side of \(4\) to see how the expression behaves.

1. Apply \(\sqrt{u^2}=|u|\): \(T(x)=|x-4|-x\). 2. When \(x\ge4\), \(x-4\ge0\), so \(|x-4|=x-4\). Then \(T(x)=(x-4)-x=-4\). 3. When \(x<4\), \(x-4<0\), so \(|x-4|=4-x\). Then \(T(x)=(4-x)-x=4-2x\). 4. The expression inside the absolute value changes sign at \(x=4\), so the rule for removing the absolute-value symbols changes there.

a) \(T(x)=-4\) b) \(T(x)=4-2x\) c) Since \(\sqrt{(x-4)^2}=|x-4|\), the expression changes form when \(x-4\) changes sign at \(x=4\).

- Use the quotient property to separate the numerator and denominator under the radical. - Apply \(\sqrt{u^2}=|u|\). - Consider the numerator and denominator of the resulting fraction separately. - Be careful when canceling a factor against an absolute value.

1. For \(k\ne2\), \(\sqrt{\frac{5}{(k-2)^2}}=\frac{\sqrt{5}}{\sqrt{(k-2)^2}}=\frac{\sqrt{5}}{|k-2|}\). 2. Thus, \(Q(k)=\frac{(k-2)\sqrt{5}}{|k-2|}\). 3. When \(k>2\), \(|k-2|=k-2\), so \(Q(k)=\sqrt{5}\). 4. When \(k<2\), \(|k-2|=-(k-2)\), so \(Q(k)=-\sqrt{5}\). 5. When \(k=2\), the denominator \((k-2)^2\) is zero, so the original expression is undefined.

For \(k>2\), \(Q(k)=\sqrt{5}\). For \(k<2\), \(Q(k)=-\sqrt{5}\). For \(k=2\), \(Q(k)\) is undefined.

- Match the power applied to an outside factor with the radical index. - Apply the power to every factor in the outside product. - Test whether the proposed transformation preserves the original value.

1. \(3y^2\sqrt[3]{\frac{2}{9y^4}}=\sqrt[3]{(3y^2)^3\cdot\frac{2}{9y^4}}=\sqrt[3]{6y^2}\). 2. For \(x>0\), \(x\sqrt[4]{2}=\sqrt[4]{x^4}\sqrt[4]{2}=\sqrt[4]{2x^4}\). 3. Noura confused multiplication with exponentiation. To move a factor inside an \(n\)th root, raise it to the \(n\)th power.

a) \(\sqrt[3]{6y^2}\) b) \(\sqrt[4]{2x^4}\); the factor must be raised to the fourth power, not multiplied by \(4\).

- Simplify each radical before deciding whether the terms are like radicals. - Make denominators inside radicals into perfect powers when needed. - Add or subtract only the coefficients of like radical terms.

1. \(\sqrt[4]{48}=\sqrt[4]{16\cdot3}=2\sqrt[4]{3}\), so \(2\sqrt[4]{48}=4\sqrt[4]{3}\). Also, \(\sqrt[4]{243}=\sqrt[4]{81\cdot3}=3\sqrt[4]{3}\). The difference is \(\sqrt[4]{3}\). 2. \(\sqrt[3]{\frac29}=\frac13\sqrt[3]{6}\), and \(\sqrt[3]{0.75}=\sqrt[3]{\frac34}=\frac12\sqrt[3]{6}\). Therefore the sum is \(\left(\frac13+\frac12\right)\sqrt[3]{6}=\frac56\sqrt[3]{6}\).

a) \(\sqrt[4]{3}\) b) \(\frac56\sqrt[3]{6}\)

- Combine the two fractions inside the radical. - Factor the numerator of the resulting fraction. - Extract perfect cubes that cancel with the outside coefficient.

1. Use the common denominator \(27u^3\): \(\frac{v^7}{9u^2}-\frac{v^6}{27u^3}=\frac{3uv^7-v^6}{27u^3}=\frac{v^6(3uv-1)}{27u^3}\). 2. Extract perfect cubes: \(\sqrt[3]{\frac{v^6(3uv-1)}{27u^3}}=\frac{v^2}{3u}\sqrt[3]{3uv-1}\). 3. The outside factors cancel, leaving \(\sqrt[3]{3uv-1}\).

\(\sqrt[3]{3uv-1}\)

- Factor each radicand. - Extract any perfect cubes. - Compare the remaining cube-root parts.

1. \(T_1=\sqrt[3]{k^3(k-1)}=k\sqrt[3]{k-1}\). 2. \(T_2=\sqrt[3]{125(k-1)}=5\sqrt[3]{k-1}\). 3. \(T_3=\sqrt[3]{(k-1)(k+1)}\), which has a different simplified radicand. 4. Therefore only \(T_1\) and \(T_2\) are like radical terms.

Only \(T_1\) and \(T_2\) are like radical terms.

- Extract powers whose exponents are multiples of the radical index. - Make a denominator inside a radical into a perfect \(n\)th power. - Compare the remaining radical parts.

1. \(\sqrt[3]{x^4y^5}=xy\sqrt[3]{xy^2}\), and \(\sqrt[3]{xy^8}=y^2\sqrt[3]{xy^2}\). The pair has the same radical part, so the terms are like terms. 2. \(\sqrt[n]{a^{2n+3}b}=a^2\sqrt[n]{a^3b}\). Also, \(\sqrt[n]{\frac{a^2}{b^{n-1}}}=\frac1b\sqrt[n]{a^2b}\). The radical parts differ, so the terms are not like terms.

a) The terms are like radical terms. b) The terms are not like radical terms.

- Move a positive coefficient under the radical by squaring it. - For two positive expressions, compare their squares. - Expand a squared sum of radicals with a binomial identity. - Compare the remaining radicals using nearby perfect squares.

1. In a), \(2\sqrt{5}=\sqrt{20}\). Since \(20>19\), \(2\sqrt{5}>\sqrt{19}\). 2. In b), both values are positive, so compare their squares. \((\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}=5+\sqrt{24}\), while \((\sqrt{10})^2=10=5+\sqrt{25}\). 3. Since \(\sqrt{24}<\sqrt{25}\), \((\sqrt{2}+\sqrt{3})^2<10\). Therefore, \(\sqrt{2}+\sqrt{3}<\sqrt{10}\).

a) \(2\sqrt{5}>\sqrt{19}\) b) \(\sqrt{10}>\sqrt{2}+\sqrt{3}\)

- Factor each radicand as a perfect square times another factor. - Combine only radicals with the same radicand. - Use the thousandths digit to round to the nearest hundredth.

1. For a), \(4\sqrt{18}-2\sqrt{50}+\sqrt{8}=12\sqrt{2}-10\sqrt{2}+2\sqrt{2}=4\sqrt{2}\approx5.66\). 2. For b), \(\frac{1}{3}\sqrt{27}+\sqrt{12}-0.5\sqrt{48}=\sqrt{3}+2\sqrt{3}-2\sqrt{3}=\sqrt{3}\approx1.73\).

a) \(4\sqrt{2}\approx5.66\) b) \(\sqrt{3}\approx1.73\)

- Look for the largest perfect-square factor in each radicand. - Does multiplying an irrational number by a nonzero rational number make it rational? - In part c), rewrite the decimal as a fraction before combining radicals.

1. \(T=5\sqrt{6}-3\sqrt{6}+2\sqrt{6}=4\sqrt{6}\). 2. Because \(\sqrt{6}\) is irrational, its nonzero rational multiple \(4\sqrt{6}\) is irrational. 3. \(2\sqrt{0.5}=2\sqrt{\frac{1}{2}}=\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\). Their sum is \(4\sqrt{2}\).

a) \(4\sqrt{6}\). b) Irrational, because \(\sqrt{6}\) is irrational and multiplying it by \(4\) does not make it rational. c) \(2\sqrt{0.5}=2\sqrt{\frac{1}{2}}=\sqrt{2}\), so \(2\sqrt{0.5}+\sqrt{18}=\sqrt{2}+3\sqrt{2}=4\sqrt{2}\).

- Move an outside factor inside by raising it to the radical index. - Factor common perfect powers from a sum. - Rewrite numerical factors as powers matching the radical index.

1. \(2y\sqrt[4]{\frac3{8y^3}}=\sqrt[4]{16y^4\cdot\frac3{8y^3}}=\sqrt[4]{6y}\). 2. \(\sqrt{18a^5+27a^4}=\sqrt{9a^4(2a+3)}=3a^2\sqrt{2a+3}\). 3. \(\sqrt[6]{64x^2}=\sqrt[6]{2^6x^2}=2x^{\frac13}=2\sqrt[3]{x}\).

a) \(\sqrt[4]{6y}\) b) \(3a^2\sqrt{2a+3}\) c) \(2\sqrt[3]{x}\)

- Factor a perfect square from each radicand where possible. - Combine radical terms only once their radicands match. - Rewrite decimals and fractions in useful forms before simplifying roots.

1. For 1), \(15\sqrt{3}-6\sqrt{3}+6\sqrt{3}=15\sqrt{3}\). 2. For 2), \(5\sqrt{2}+4\sqrt{2}-9\sqrt{2}=0\). 3. For 3), rewrite \(\frac{1}{5}=\frac{5}{25}\) before taking the square root. Then \(2\sqrt{5}-\frac{3\sqrt{5}}{10}+\frac{\sqrt{5}}{5}=\frac{19\sqrt{5}}{10}\).

1) \(15\sqrt{3}\). 2) \(0\). 3) \(\frac{19\sqrt{5}}{10}\).

- Simplify each radical separately before comparing or combining. - Distribute the subtraction across the second set of parentheses. - Do not add radicands unless a valid radical property applies. - A radical expression can simplify to a rational number.

1. \(\sqrt{12}+\sqrt{48}=2\sqrt{3}+4\sqrt{3}=6\sqrt{3}\), and \(\sqrt{108}=6\sqrt{3}\). Therefore, the equation is true. 2. \((3\sqrt{20}-\sqrt{45})-(\sqrt{80}-\sqrt{5})=(6\sqrt{5}-3\sqrt{5})-(4\sqrt{5}-\sqrt{5})=3\sqrt{5}-3\sqrt{5}=0\). 3. The result \(0\) is rational.

a) True; both sides equal \(6\sqrt{3}\). b) The value is \(0\), which is rational.

- Factor out square factors before comparing the remaining radicals. - What changes when two radicands are added or subtracted rather than multiplied? - In part c), can you rewrite the fraction inside the square root to reveal the same radical part in both terms?

1. In a), \(\sqrt{20x^3}=2x\sqrt{5x}\) and \(\sqrt{45x}=3\sqrt{5x}\). Both have the same remaining radicand, \(5x\). 2. In b), the radicands \(a^2+b^2\) and \(a^2-b^2\) are generally different, so these are not like radical terms in general. They can match for special values such as \(b=0\). 3. In c), since \(y>0\), \(\sqrt{\frac{1}{3y}}=\sqrt{\frac{3y}{(3y)^2}}=\frac{\sqrt{3y}}{3y}\). Both terms have the same radical part, \(\sqrt{3y}\); no denominator rationalization is needed.

a) Like radical terms. b) Not like radical terms in general; special values can make the expressions equal. c) Like radical terms.

- Factor a difference of squares in part a. - Factor the radicand of the first term in part b. - Extract a perfect cube from the numerator or denominator.

1. In part a, \(x^2-y^2=(x-y)(x+y)\), so \(A=\sqrt{x+y}=B\). The terms are identical. 2. In part b, \(A=\sqrt[3]{a^3(a-b)}=a\sqrt[3]{a-b}\). Also, \(B=\frac1a\sqrt[3]{a-b}\). They have the same cube-root part, so they are like radical terms.

a) The terms are identical and therefore like radical terms. b) The terms are like radical terms with radical part \(\sqrt[3]{a-b}\).

- Split exponents into multiples of the radical index plus a remainder. - Extract the factors with divisible exponents. - Treat numerator and denominator separately when useful.

1. \(\sqrt[n]{x^{3n}y^{n+2}}=x^3y\sqrt[n]{y^2}\). 2. \(\sqrt[k]{a^{2k+1}b^{k-1}}=a^2\sqrt[k]{ab^{k-1}}\). 3. \(\sqrt[n]{\frac{p^{n+3}}{q^{2n}}}=\frac{p}{q^2}\sqrt[n]{p^3}\).

1) \(x^3y\sqrt[n]{y^2}\) 2) \(a^2\sqrt[k]{ab^{k-1}}\) 3) \(\frac{p}{q^2}\sqrt[n]{p^3}\)

- Separate powers whose exponents match the radical index. - Simplify the left side of each equation independently. - Use the quotient property for radicals in part 2.

1. \(\sqrt[n]{a^{2n+1}b^{n+2}}=\sqrt[n]{a^{2n}ab^nb^2}=a^2b\sqrt[n]{ab^2}\). The transformation is correct. 2. \(\sqrt[k+1]{\frac{x^{k+2}}{y^{k+1}}}=\frac{\sqrt[k+1]{x^{k+1}x}}{\sqrt[k+1]{y^{k+1}}}=\frac{x}{y}\sqrt[k+1]{x}\). The transformation is correct.

1) Correct 2) Correct

- Extract perfect cubes and perfect squares. - Distribute the negative sign across the second parentheses. - Combine only terms with the same radical part.

1. \(\sqrt[3]{27x}=3\sqrt[3]{x}\), \(\sqrt[3]{8x}=2\sqrt[3]{x}\), \(\sqrt{64y}=8\sqrt y\), and \(\sqrt{4y}=2\sqrt y\). 2. Substitute and distribute the subtraction: \(3\sqrt[3]{x}+8\sqrt y-4\sqrt[3]{x}+2\sqrt y\). 3. Combine like radical terms to get \(-\sqrt[3]{x}+10\sqrt y\).

\(-\sqrt[3]{x}+10\sqrt y\)

- Distribute the factor outside the parentheses. - Combine cube roots in products and quotients. - Look for perfect cubes after combining radicands.

1. Distribute the first cube root: \(\sqrt[3]{9}\sqrt[3]{3}+\sqrt[3]{9}\sqrt[3]{24}=\sqrt[3]{27}+\sqrt[3]{216}=3+6\). 2. Simplify the quotient: \(\frac{\sqrt[3]{375}}{\sqrt[3]{3}}=\sqrt[3]{125}=5\). 3. Therefore the expression equals \(3+6-5=4\), which is a natural number.

\(4\)

- Simplify every radical first. - A minus sign before parentheses changes the signs inside. - Combine terms only when their radicands match.

1. Rewrite the radicals: \(\sqrt{45}=3\sqrt{5}\), \(1.5\sqrt{12}=3\sqrt{3}\), \(2\sqrt{20}=4\sqrt{5}\), \(\sqrt{48}=4\sqrt{3}\), and \(\sqrt{27}=3\sqrt{3}\). 2. Remove parentheses carefully: \(3\sqrt{5}-3\sqrt{3}-4\sqrt{5}+4\sqrt{3}+3\sqrt{3}\). 3. Combine like radicals to obtain \(4\sqrt{3}-\sqrt{5}\).

\(4\sqrt{3}-\sqrt{5}\)

- Rewrite each radicand using a squared-binomial pattern. - Use absolute value when taking the square root of a squared expression. - Determine the sign of each expression inside the absolute-value symbols on the given interval. - Recall how \(|u|\) simplifies when \(u\) is negative.

1. Rewrite each radicand as a perfect square: \(4x^2-12x+9=(2x-3)^2\) and \(4x^2+12x+9=(2x+3)^2\). 2. Therefore, \(A=|2x-3|+|2x+3|\). 3. On the interval \(-1.5\le x\le1.5\), \(2x-3\le0\), so \(|2x-3|=3-2x\). Also, \(2x+3\ge0\), so \(|2x+3|=2x+3\). 4. Thus, \(A=(3-2x)+(2x+3)=6\).

\(6\)

- Factor the common factor from each radicand. - Look for squared-binomial patterns. - Use the conditions on \(a\) and \(b\) to remove the absolute-value symbols correctly. - Pay close attention to signs when subtracting the second expression.

1. Factor \(c\) from each radicand: \(T=\sqrt{c(a^2-2ab+b^2)}-\sqrt{c(a^2+2ab+b^2)}\). 2. Use the squared-binomial patterns: \(T=\sqrt{c(a-b)^2}-\sqrt{c(a+b)^2}\). 3. Extract the squared factors: \(T=\sqrt{c}|a-b|-\sqrt{c}|a+b|\). 4. Since \(a>b>0\), both \(a-b\) and \(a+b\) are positive. Therefore, \(T=\sqrt{c}(a-b)-\sqrt{c}(a+b)\). 5. Combine like terms: \(T=\sqrt{c}(a-b-a-b)=-2b\sqrt{c}\).

\(-2b\sqrt{c}\)

- Identify the binomial-square and difference-of-squares patterns. - Square each radical carefully. - Which expansion keeps a cross term, and which cancels it?

1. Using the square of a binomial, \(A=7+2\sqrt{14}+2=9+2\sqrt{14}\). 2. Using the difference of squares, \(B=7-2=5\). 3. In \(A\), the middle term \(2\sqrt{14}\) remains and is irrational. In \(B\), the two cross terms cancel, leaving only the difference of the squared radicals, which is rational.

a) \(A=9+2\sqrt{14}\) and \(B=5\) b) The cross term remains in \(A\), but the conjugate factors in \(B\) cancel the cross terms and leave \(7-2\).

- Look for perfect-square factors inside each radical. - Can you extract a decimal perfect-square factor from the decimal radicand? - Combine coefficients only after finding like radical terms.

1. Simplify each root: \(\sqrt{45}=3\sqrt{5}\), \(\sqrt{20}=2\sqrt{5}\), \(\sqrt{125}=5\sqrt{5}\), \(\sqrt{80}=4\sqrt{5}\), \(\sqrt{0.2}=\sqrt{0.04\cdot5}=\frac{\sqrt{5}}{5}\), and \(\sqrt{180}=6\sqrt{5}\). 2. Combine: \(\left(3-2+5-4+\frac{1}{5}+1-6\right)\sqrt{5}=-\frac{14\sqrt{5}}{5}\).

\(-\frac{14\sqrt{5}}{5}\).

- Group terms by the radical part that remains after simplification. - Convert the mixed number to an improper fraction. - Check subtraction signs when combining like radicals.

1. The terms with \(\sqrt{2}\) become \(3\sqrt{2}+\frac{\sqrt{2}}{2}+10\sqrt{2}=\frac{27}{2}\sqrt{2}\). 2. The terms with \(\sqrt{3}\) become \(-2\sqrt{3}-\frac{4\sqrt{3}}{3}-6\sqrt{3}=-\frac{28}{3}\sqrt{3}\). For the mixed number, \(\sqrt{5\frac{1}{3}}=\sqrt{\frac{16}{3}}=\sqrt{\frac{48}{9}}=\frac{4\sqrt{3}}{3}\). 3. The sum is \(\frac{27}{2}\sqrt{2}-\frac{28}{3}\sqrt{3}\).

\(\frac{27}{2}\sqrt{2}-\frac{28}{3}\sqrt{3}\).

- Extract perfect-square factors from the radicals. - Use the relationship between the square root of a square and absolute value. - Use \(x>y\) to determine the sign of \(x-y\). - Combine terms with the same radical factor.

1. Since \(x>0\) and \(z>0\), \(\sqrt{9x^2z}=3x\sqrt{z}\) and \(\sqrt{4x^2z}=2x\sqrt{z}\). 2. Since \(x>y>0\), both \(x-y\) and \(x+y\) are positive. Thus, \(\sqrt{(x-y)^2}=x-y\) and \(\sqrt{(x+y)^2}=x+y\). 3. Substitute these results: \(x\sqrt{z}+3x\sqrt{z}-2x\sqrt{z}+(x-y)-(x+y)\). 4. Combine like terms: \((1+3-2)x\sqrt{z}=2x\sqrt{z}\) and \(x-y-x-y=-2y\). 5. The simplified expression is \(2x\sqrt{z}-2y\).

\(2x\sqrt{z}-2y\)

- Factor a common factor from the first radicand and look for a squared-binomial pattern. - Remember that the square root of a square is an absolute value. - Use \(a>b>0\) to determine the signs of the expressions inside the absolute values.

1. Factor the first radicand: \(4a^2-8ab+4b^2=4(a^2-2ab+b^2)=4(a-b)^2\). 2. Since \(a>b\), \(a-b>0\), so \(\sqrt{4(a-b)^2}=2(a-b)\). 3. Since \(a>0\) and \(a+b>0\), \(\sqrt{a^2}=a\) and \(\sqrt{(a+b)^2}=a+b\). 4. Therefore, \(T=2(a-b)+a-(a+b)\). 5. Combine like terms: \(T=2a-2b+a-a-b=2a-3b\).

\(2a-3b\)

- Identify the expressions that play the roles of \(u\) and \(v\) in \((u-v)(u+v)\). - Square every factor in a product such as \(2\sqrt{3}\). - Use parentheses when subtracting an entire squared expression.

1. Each product is a product of conjugates, so use \((u-v)(u+v)=u^2-v^2\). 2. For 1), \(5^2-(2\sqrt{3})^2=25-12=13\). 3. For 2), \((x\sqrt{y})^2-(y\sqrt{x})^2=x^2y-xy^2=xy(x-y)\). 4. For 3), \((2\sqrt{a+b})^2-(\sqrt{a-b})^2=4(a+b)-(a-b)=3a+5b\).

1) \(13\) 2) \(xy(x-y)\) 3) \(3a+5b\)

- Identify a difference-of-cubes identity. - Let the two simple cube roots represent new variables. - Match the three terms in the first factor to \(u^2\), \(uv\), and \(v^2\).

1. Let \(u=\sqrt[3]{7}\) and \(v=\sqrt[3]{2}\). Then the expression is \((u^2+uv+v^2)(u-v)\). 2. Use \((u^2+uv+v^2)(u-v)=u^3-v^3\). 3. Therefore \(A=7-2=5\).

\(A=5\)

- Extract perfect-square factors from \(18\), \(50\), and \(8\). - Rewrite every radical term with the same radicand. - Combine the coefficients of the like radical terms.

1. Simplify each radical term: \(2\sqrt{18x}=6\sqrt{2x}\), \(\sqrt{50x}=5\sqrt{2x}\), and \(3\sqrt{8x}=6\sqrt{2x}\). 2. Substitute these forms: \(T=(6\sqrt{2x}-5\sqrt{2x}+4)+(5\sqrt{2x}-6\sqrt{2x})\). 3. The radical coefficients add to \(6-5+5-6=0\), so all radical terms cancel and \(T=4\).

\(4\)

- Extract the largest perfect-square factor from each radicand. - Rewrite all three radicals with a common radicand. - Compare the fully simplified expressions on the two sides.

1. Simplify the left side: \(\sqrt{27y}=3\sqrt{3y}\) and \(\sqrt{48y}=4\sqrt{3y}\), so \(\sqrt{27y}+\sqrt{48y}=7\sqrt{3y}\). 2. Simplify the right side: \(\sqrt{147y}=7\sqrt{3y}\). 3. Both sides simplify to the same expression for every \(y\ge0\), so the equation is true on the stated domain.

Yes. Both sides simplify to \(7\sqrt{3y}\), so the equation is true for every \(y\ge0\).

- Rewrite every radical as a rational exponent. - Work from the innermost radical outward. - Multiply exponents for a power raised to another power.

1. \(\sqrt[3]{\sqrt b}=b^{\frac12\cdot\frac13}=b^{\frac16}=\sqrt[6]{b}\). 2. \(\sqrt{c\sqrt[3]{c}}=(c\cdot c^{\frac13})^{\frac12}=c^{\frac23}=\sqrt[3]{c^2}\). 3. \(\sqrt[5]{d^2\sqrt d}=(d^{2+\frac12})^{\frac15}=d^{\frac12}=\sqrt d\).

1) \(\sqrt[6]{b}\) 2) \(\sqrt[3]{c^2}\) 3) \(\sqrt d\)

- Combine radicals with the same index in a quotient. - Subtract variable exponents after combining the radicands. - Simplify the numerical coefficient outside the radical separately.

1. \(\frac{\sqrt[3]{250}}{\sqrt[3]{2}}=\sqrt[3]{125}=5\). 2. \(\frac{\sqrt[3]{54a^7}}{\sqrt[3]{2a}}=\sqrt[3]{27a^6}=3a^2\). 3. \(\frac{10\sqrt{0.2}}{2\sqrt5}=5\sqrt{\frac{0.2}{5}}=5\sqrt{0.04}=1\).

1) \(5\) 2) \(3a^2\) 3) \(1\)

- Rewrite a positive number as a product involving its square root when that creates a common factor. - Extract perfect-square variable factors from radicals. - Look for a common radical factor in a sum or difference. - In a fraction, check whether the numerator and denominator share a radical factor.

1. Write \(5=\sqrt{5}\cdot\sqrt{5}\), then factor: \(5+\sqrt{5}=\sqrt{5}(\sqrt{5}+1)\). 2. Since \(x\ge0\), \(\sqrt{x^3}=x\sqrt{x}\). Thus, \(x\sqrt{x}+\sqrt{x}=\sqrt{x}(x+1)\). 3. Factor out \(\sqrt{12}\): \(\sqrt{24}-\sqrt{12}=\sqrt{12}(\sqrt{2}-1)=2\sqrt{3}(\sqrt{2}-1)\). 4. Factor \(\sqrt{5}\) from the numerator: \(\frac{\sqrt{5}(\sqrt{6}+2)}{\sqrt{5}}=\sqrt{6}+2\).

1) \(\sqrt{5}(\sqrt{5}+1)\) 2) \(\sqrt{x}(x+1)\) 3) \(2\sqrt{3}(\sqrt{2}-1)\) 4) \(\sqrt{6}+2\)

- Replace division by a radical with multiplication by its reciprocal. - Distribute the reciprocal radical across the numerator. - Combine products of radicals and simplify their radicands. - Use the fact that all variables are positive when taking square roots of squares.

1. Dividing by \(\sqrt{1/x}\) is equivalent to multiplying by \(\sqrt{x}\). Therefore, \((\sqrt{x^3}+\sqrt{x})\sqrt{x}=\sqrt{x^4}+\sqrt{x^2}=x^2+x\). 2. Dividing by \(\sqrt{b/a^2}\) is equivalent to multiplying by \(\sqrt{a^2/b}\). Thus, \((\sqrt{a^2b}-\sqrt{b^3})\sqrt{\frac{a^2}{b}}=\sqrt{a^4}-\sqrt{a^2b^2}=a^2-ab\).

1) \(x^2+x\) 2) \(a^2-ab\)

- Look for a difference of squares inside or outside a radical. - Rewrite cube roots with exponent \(\frac13\) if that reveals a familiar identity. - Identify the greatest common factor in part 3.

1. Since \(a^2-4=(a-2)(a+2)\), \(\sqrt{a^2-4}-\sqrt{a+2}=\sqrt{a+2}(\sqrt{a-2}-1)\). 2. View the expression as a difference of squares: \((\sqrt[3]{x})^2-(\sqrt[3]{y})^2=(\sqrt[3]{x}-\sqrt[3]{y})(\sqrt[3]{x}+\sqrt[3]{y})\). 3. Factor out \(z\): \(z\sqrt z-z=z(\sqrt z-1)\).

1) \(\sqrt{a+2}(\sqrt{a-2}-1)\) 2) \((\sqrt[3]{x}-\sqrt[3]{y})(\sqrt[3]{x}+\sqrt[3]{y})\) 3) \(z(\sqrt z-1)\)

- Use the quotient property for square roots. - Simplify each radicand by extracting a factor of \(7\). - Division by a common nonzero factor distributes across a sum or difference. - Check what happens when the same nonzero radical factor appears in the numerator and denominator.

1. Distributing the division gives \(6\sqrt{4}-2\sqrt{9}+\sqrt{16}=12-6+4=10\). 2. Alternatively, \(\sqrt{28}=2\sqrt{7}\), \(\sqrt{63}=3\sqrt{7}\), and \(\sqrt{112}=4\sqrt{7}\). The numerator becomes \(12\sqrt{7}-6\sqrt{7}+4\sqrt{7}=10\sqrt{7}\), and \(10\sqrt{7}\div\sqrt{7}=10\). 3. Both methods apply equivalent algebraic and radical properties to the same expression, so they preserve its value.

a) \(10\) b) \(10\) c) Both methods use valid equivalent transformations of the same expression.

- Divide every numerator term by the denominator. - Combine cube roots in each quotient. - Evaluate the resulting perfect cubes.

1. Divide each term in the numerator by \(\sqrt[3]{2}\): \(\frac12\sqrt[3]{8}+\frac13\sqrt[3]{27}-\sqrt[3]{1}\). 2. Evaluate the cube roots: \(\frac12\cdot2+\frac13\cdot3-1=1\). 3. The result \(1\) is a natural number.

\(1\), which is a natural number.

- Rewrite the numerator to expose a factor that also appears in the denominator. - Express a nonnegative variable as a product of two equal square roots. - Use the product property to rewrite \(\sqrt{15}\). - Confirm that any canceled factor is nonzero.

1. Factor \(\sqrt{7}\) from the numerator: \(\frac{\sqrt{7}(1+\sqrt{7})}{\sqrt{7}}=1+\sqrt{7}\). 2. Since \(a=\sqrt{a}\cdot\sqrt{a}\), factor the numerator: \(\frac{\sqrt{a}(\sqrt{a}-1)}{\sqrt{a}-1}=\sqrt{a}\). The condition \(a\ne1\) ensures the canceled factor is nonzero. 3. Rewrite \(\sqrt{15}=\sqrt{5}\sqrt{3}\), then factor and cancel: \(\frac{\sqrt{5}(\sqrt{3}-1)}{\sqrt{5}}=\sqrt{3}-1\).

1) \(1+\sqrt{7}\) 2) \(\sqrt{a}\) 3) \(\sqrt{3}-1\)

- Group terms in pairs that have a common radical factor. - After factoring each group, look for a common binomial. - Rewrite a difference of nonnegative variables as a difference of squares involving their square roots. - Treat \(x\) and \(y\) as \((\sqrt{x})^2\) and \((\sqrt{y})^2\).

1. Group the first two and last two terms: \((\sqrt{uv}-\sqrt{3u})+(\sqrt{2v}-\sqrt{6})\). Factor each group to get \(\sqrt{u}(\sqrt{v}-\sqrt{3})+\sqrt{2}(\sqrt{v}-\sqrt{3})\). Factoring the common binomial gives \((\sqrt{u}+\sqrt{2})(\sqrt{v}-\sqrt{3})\). 2. Rewrite \(x-y=(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})\), and rewrite \(-3\sqrt{x}+3\sqrt{y}=-3(\sqrt{x}-\sqrt{y})\). Factoring the common binomial gives \((\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}-3)\).

1) \((\sqrt{u}+\sqrt{2})(\sqrt{v}-\sqrt{3})\) 2) \((\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}-3)\)

- Factor expressions under the radicals when possible. - Rewrite a nonnegative expression without a radical as the square of its square root. - In part 2, recognize a perfect-square trinomial under the first radical.

1. Factor \(x^2-16=(x-4)(x+4)\), and write \(x-4=(\sqrt{x-4})^2\). Then \(\sqrt{x^2-16}+x-4=\sqrt{x-4}\sqrt{x+4}+(\sqrt{x-4})^2\), so the factored form is \(\sqrt{x-4}(\sqrt{x+4}+\sqrt{x-4})\). 2. Since \(x^2-2xy+y^2=(x-y)^2\) and \(x-y\ge0\), the first term is \(x-y\). Thus, \((x-y)+\sqrt{x-y}=\sqrt{x-y}(\sqrt{x-y}+1)\).

1) \(\sqrt{x-4}(\sqrt{x+4}+\sqrt{x-4})\) 2) \(\sqrt{x-y}(\sqrt{x-y}+1)\)

- Rewrite radicals as rational exponents. - Apply the outer power to the coefficient and radical factor. - Separate any whole-number part of the resulting exponent.

1. \((\sqrt[4]{a^3})^2=a^{\frac32}=a\sqrt a\). 2. \((-\sqrt[3]{b^2})^4=b^{\frac83}=b^2\sqrt[3]{b^2}\). 3. \(\left(-\frac23\sqrt[5]{x^3}\right)^3=-\frac8{27}x^{\frac95}=-\frac8{27}x\sqrt[5]{x^4}\).

a) \(a\sqrt a\) b) \(b^2\sqrt[3]{b^2}\) c) \(-\frac8{27}x\sqrt[5]{x^4}\)

- Expand the squared binomial first. - Factor a large perfect square from \(1728\). - Check whether the radical terms cancel.

1. Expanding gives \((4-3\sqrt{3})^2=16-24\sqrt{3}+27=43-24\sqrt{3}\). 2. Since \(1728=576\cdot3\), \(\sqrt{1728}=24\sqrt{3}\). 3. The radical terms cancel: \(43-24\sqrt{3}+24\sqrt{3}=43\).

\(43\)

- How does squaring change the index of a fourth root? - Can you rewrite the radicand so that it contains a perfect-square factor? - How do the positive assumptions on \(a\) and \(b\) affect square-root simplification?

1. Square the factors: \(A=\frac{a^2}{9b^2}\sqrt{\frac{27b^3}{a}}\). 2. Rewrite the radicand as \(\frac{27b^3}{a}=\left(\frac{3b}{a}\right)^2(3ab)\). 3. Since \(a>0\) and \(b>0\), \(\sqrt{\frac{27b^3}{a}}=\frac{3b}{a}\sqrt{3ab}\). 4. Therefore \(A=\frac{a^2}{9b^2}\cdot\frac{3b}{a}\sqrt{3ab}=\frac{a\sqrt{3ab}}{3b}=B\).

\(A=\frac{a^2}{9b^2}\sqrt{\frac{27b^3}{a}}=\frac{a^2}{9b^2}\cdot\frac{3b\sqrt{3ab}}{a}=\frac{a\sqrt{3ab}}{3b}=B\). Yes, they are equal.

- Apply the square-of-a-binomial identities before simplifying. - Square both the coefficient and the radical in each first and last term. - In c), combine the product \(\sqrt{xy}\sqrt{x}\) under one radical and extract the perfect-square factor.

1. In a), use \((u-v)^2=u^2-2uv+v^2\): \((5\sqrt{z})^2-2(5\sqrt{z})(1)+1^2=25z-10\sqrt{z}+1\). 2. In b), use \((u+v)^2=u^2+2uv+v^2\): \(\left(\frac{1}{2}\sqrt{x}\right)^2+2\left(\frac{1}{2}\sqrt{x}\right)(4\sqrt{y})+(4\sqrt{y})^2=\frac{x}{4}+4\sqrt{xy}+16y\). 3. In c), expand: \((\sqrt{xy})^2-2(\sqrt{xy})(3\sqrt{x})+(3\sqrt{x})^2\). Because \(x>0\), \(\sqrt{xy}\sqrt{x}=x\sqrt{y}\). The result is \(xy-6x\sqrt{y}+9x\).

a) \(25z-10\sqrt{z}+1\) b) \(\frac{x}{4}+4\sqrt{xy}+16y\) c) \(xy-6x\sqrt{y}+9x\)

- Combine like radicals as you would like terms. - Simplify \(\sqrt{32}\) before squaring. - Use \((x+y)^2=x^2+2xy+y^2\). - Test a simple numerical case before accepting a statement as universally true.

1. Combining first gives \((3\sqrt{7})^2=63\). Expanding instead gives \(7+2(\sqrt{7})(2\sqrt{7})+28=7+28+28=63\). 2. Since \(\sqrt{32}=4\sqrt{2}\), \((\sqrt{32}-3\sqrt{2})^2=(\sqrt{2})^2=2\). 3. For example, with \(a=b=1\), \((\sqrt{1}+\sqrt{1})^2=4\), while \(a+b=2\). The claim omits the cross term \(2\sqrt{ab}\).

a) \(63\) b) \(2\) c) The claim is false; the missing term is \(2\sqrt{ab}\).

- Extract perfect-square factors before squaring. - Apply the square-of-a-difference formula carefully. - To verify \(\sqrt{A}=B\), check both \(B^2=A\) and \(B\ge0\).

1. Since \(\sqrt{12}=2\sqrt{3}\), \(\left(\frac{1}{2}\sqrt{12}+\sqrt{3}\right)^2=(2\sqrt{3})^2=12\). 2. \((3\sqrt{5}-2\sqrt{3})^2=45-12\sqrt{15}+12=57-12\sqrt{15}\). 3. \((2+\sqrt{3})^2=4+4\sqrt{3}+3=7+4\sqrt{3}\). Also, \(2+\sqrt{3}>0\), so it is the principal square root of \(7+4\sqrt{3}\).

a) \(12\) b) \(57-12\sqrt{15}\) c) \((2+\sqrt{3})^2=7+4\sqrt{3}\) and \(2+\sqrt{3}>0\).

- Expand each squared binomial. - Recall that squaring a square root returns its nonnegative radicand. - Simplify the product of conjugate expressions using a difference of squares. - Look for radical terms that cancel.

1. For a), expanding gives \(9+\sqrt{17}+9-\sqrt{17}+2\sqrt{(9+\sqrt{17})(9-\sqrt{17})}\). 2. The product is \(81-17=64\), so the expression equals \(18+2\sqrt{64}=18+16=34\). 3. For b), expanding gives \(10-2\sqrt{21}+10+2\sqrt{21}-2\sqrt{(10-2\sqrt{21})(10+2\sqrt{21})}\). 4. The product is \(100-84=16\), so the expression equals \(20-2\sqrt{16}=20-8=12\).

a) \(34\) b) \(12\)

- Expand the large squared sum first. - Use a difference of squares for the product inside the cross term. - Recall the definition of a perfect square. - Check whether the final value is the square of an integer.

1. Expanding the squared sum gives \(11+\sqrt{72}+11-\sqrt{72}+2\sqrt{(11+\sqrt{72})(11-\sqrt{72})}\). 2. The conjugate product is \(121-72=49\), so the squared sum equals \(22+2\sqrt{49}=22+14=36\). 3. Therefore, \(T=\frac{1}{4}\cdot36=9=3^2\), which is a perfect square.

\(T=9=3^2\), so the value is a perfect square.

- Simplify inside the parentheses before expanding when possible. - For three terms, include every square and each pairwise cross product twice. - Combine like radicals before squaring. - Simplify each radical enough to identify all like radical terms.

1. For a), squaring the three-term sum gives \(7+3+4+2\sqrt{21}+4\sqrt{7}+4\sqrt{3}\). Therefore, the result is \(14+2\sqrt{21}+4\sqrt{7}+4\sqrt{3}\). 2. For b), \(\sqrt{20}=2\sqrt{5}\), so the expression inside the parentheses simplifies to \(\sqrt{2}\). Its square is \(2\).

a) \(14+2\sqrt{21}+4\sqrt{7}+4\sqrt{3}\) b) \(2\)

- Simplify the radicals before squaring. - Combine like radicals inside each set of parentheses. - Apply the square-of-a-difference formula carefully. - Check whether the radical terms cancel so the result is a natural number.

1. Since \(\sqrt{12}=2\sqrt{3}\), the first expression becomes \((3\sqrt{3}-\sqrt{2})^2=27-6\sqrt{6}+2=29-6\sqrt{6}\). 2. Since \(\sqrt{8}=2\sqrt{2}\) and \(\sqrt{32}=4\sqrt{2}\), the second expression becomes \((7\sqrt{2})^2=98\), which is a natural number.

a) \(29-6\sqrt{6}\) b) \(98\), a natural number

- Use \((a\pm b)^2=a^2\pm2ab+b^2\). - Look for two numbers whose sum and product match the nonradical and radical parts. - Try expressing the large radicand as the square of a binomial. - Verify that the simplified expression is nonnegative.

1. \((\sqrt{5}+\sqrt{2})^2=5+2\sqrt{10}+2=7+2\sqrt{10}\). 2. Since \((\sqrt{5}-\sqrt{2})^2=7-2\sqrt{10}\) and \(\sqrt{5}-\sqrt{2}>0\), \(\sqrt{7-2\sqrt{10}}=\sqrt{5}-\sqrt{2}\). 3. The numbers \(7\) and \(5\) have sum \(12\) and product \(35\). Thus, \((\sqrt{7}+\sqrt{5})^2=12+2\sqrt{35}\), so \(\sqrt{12+2\sqrt{35}}=\sqrt{7}+\sqrt{5}\).

1. \(7+2\sqrt{10}\) 2. \(\sqrt{5}-\sqrt{2}\) 3. \(\sqrt{7}+\sqrt{5}\)

- Rewrite each radical as a rational exponent. - Multiply exponents for nested powers. - Write \(729\) as a power of \(3\).

1. \(\sqrt{\sqrt{625}}=\sqrt{25}=5\), and \(\sqrt[4]{625}=5\). The values are equal. 2. \(\sqrt{\sqrt a}=(a^{\frac12})^{\frac12}=a^{\frac14}=\sqrt[4]{a}\). 3. Since \(729=3^6\), \(\sqrt[3]{\sqrt[n]{729}}=3^{\frac6{3n}}=3\). Thus \(\frac6{3n}=1\), so \(n=2\).

a) Both values are \(5\). b) \(\sqrt{\sqrt a}=a^{\frac14}=\sqrt[4]{a}\). c) \(n=2\)

- Multiply radical indices when roots are nested. - Write the numerical radicands as powers of \(2\). - Subtract exponents when dividing powers of \(x\).

1. \(A=\sqrt[6]{4096}=\sqrt[6]{2^{12}}=4\). 2. \(B=\sqrt[8]{256}\cdot2=\sqrt[8]{2^8}\cdot2=4\). Thus \(A=B\). 3. \(\sqrt[4]{\sqrt{x^{16}}}=x^2\). Therefore \(\frac{x^2}{x^{\frac12}}=x^{\frac32}=x\sqrt x\).

a) \(A=B=4\) b) \(x\sqrt x\)

- Rewrite inner radicals as rational exponents. - Combine powers with base \(2\). - For a base greater than \(1\), compare the resulting exponents.

1. \(A=(2\cdot2^{\frac12})^{\frac13}=2^{\frac12}=\sqrt2\). 2. \(B=(2\cdot2^{\frac13})^{\frac14}=2^{\frac13}=\sqrt[3]{2}\). 3. Since \(2>1\) and \(\frac12>\frac13\), \(2^{\frac12}>2^{\frac13}\). Therefore \(A>B\).

\(A>B\), because \(A=\sqrt2\) and \(B=\sqrt[3]{2}\).

- Compute \(x+y\) and \(x-y\) separately first. - Combine fractions that have the same denominator. - Dividing by a fraction is equivalent to multiplying by its reciprocal.

1. Add the given expressions: \(x+y=\frac{\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}}{\sqrt{a}}=2\). 2. Subtract them: \(x-y=\frac{\sqrt{a}+\sqrt{b}-(\sqrt{a}-\sqrt{b})}{\sqrt{a}}=\frac{2\sqrt{b}}{\sqrt{a}}\). 3. Therefore, \(T=\frac{2}{2\sqrt{b}/\sqrt{a}}=\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\).

\(T=\sqrt{\frac{a}{b}}\)

- Substitute for \(x\) inside the radicand and expand. - Look for a perfect-square expression involving \(a\) and \(1/a\). - Use \(a>1\) when taking the square root. - Simplify the numerator and denominator before dividing.

1. Substitute for \(x\): \(x^2-4=\left(a+\frac{1}{a}\right)^2-4=a^2-2+\frac{1}{a^2}=\left(a-\frac{1}{a}\right)^2\). 2. Since \(a>1\), \(a-\frac{1}{a}>0\), so \(\sqrt{x^2-4}=a-\frac{1}{a}\). 3. The denominator becomes \(a+\frac{1}{a}-\left(a-\frac{1}{a}\right)=\frac{2}{a}\). 4. Thus, \(\frac{a-1/a}{2/a}=\frac{a^2-1}{2}\).

\(\frac{a^2-1}{2}\)

- Try the easier case \(x=8\), \(y=1\) to see what the expression does. - Can you recognize a pattern if you temporarily use shorter letters for \(x^{\frac{1}{3}}\) and \(y^{\frac{1}{3}}\)? - Make sure any canceled factor really multiplies the entire numerator and denominator.

1. View the numerator as a difference of cubes: \(x-y=(x^{\frac{1}{3}})^3-(y^{\frac{1}{3}})^3\). 2. Factor it as \((x^{\frac{1}{3}}-y^{\frac{1}{3}})(x^{\frac{2}{3}}+x^{\frac{1}{3}}y^{\frac{1}{3}}+y^{\frac{2}{3}})\). 3. Cancel the common factor in the fraction to obtain \(x^{\frac{1}{3}}-y^{\frac{1}{3}}\). 4. The expression in parentheses reduces to \(x^{\frac{1}{3}}\). 5. Thus \(B=x^{\frac{1}{3}}x^{-\frac{1}{3}}=x^0=1\).

\(1\)

- Raise the outside coefficient to the fourth power before moving it inside. - Rewrite the nested radical using rational exponents. - Compare radicands once the radical indices match.

1. \(2\sqrt[4]{3}=\sqrt[4]{2^4\cdot3}=\sqrt[4]{48}\). Since \(48<50\), \(\sqrt[4]{50}\) is greater. 2. \(\sqrt[3]{x\sqrt x}=\left(x\cdot x^{\frac12}\right)^{\frac13}=x^{\frac12}=\sqrt x\).

a) \(\sqrt[4]{50}\) is greater. b) \(\sqrt x\)

- Square each positive value. - Compare the resulting rational numbers. - Squaring preserves the order of positive numbers.

1. The squares are \((2\sqrt{11})^2=44\), \((3\sqrt{5})^2=45\), and \(6.5^2=42.25\). 2. Since all three values are positive and \(42.25<44<45\), the original values satisfy \(6.5<2\sqrt{11}<3\sqrt{5}\).

\(6.5<2\sqrt{11}<3\sqrt{5}\)

- Move a positive coefficient under the radical by squaring it. - Rewrite a positive integer as the square root of its square when useful. - Compare the radicands before calculating decimal approximations.

1. In a), \(3\sqrt{11}=\sqrt{99}\), while \(10=\sqrt{100}\). Therefore, \(3\sqrt{11}<10\). Numerically, \(3\sqrt{11}\approx9.95<10\). 2. In b), \(5\sqrt{6}=\sqrt{150}\) and \(6\sqrt{5}=\sqrt{180}\). Therefore, \(5\sqrt{6}<6\sqrt{5}\). Numerically, \(5\sqrt{6}\approx12.25\) and \(6\sqrt{5}\approx13.42\).

a) \(3\sqrt{11}<10\); approximately \(9.95<10\) b) \(5\sqrt{6}<6\sqrt{5}\); approximately \(12.25<13.42\)

- Extract perfect cubes or squares first. - Compare both the radical index and the remaining radicand. - Coefficients do not have to match for terms to be like.

1. \(\sqrt[3]{24}=2\sqrt[3]{3}\), and \(\sqrt[3]{81}=3\sqrt[3]{3}\). They are like terms. 2. \(\sqrt{12}=2\sqrt3\) has index \(2\), while \(\sqrt[3]{12}\) has index \(3\). They are not like terms. 3. \(\frac35\sqrt[3]{250}=3\sqrt[3]{2}\), and \(0.5\sqrt[3]{128}=2\sqrt[3]{2}\). They are like terms.

a) Like terms b) Not like terms c) Like terms

- Combine products and quotients of square roots into a single radical. - Simplify inside the radical before taking a decimal approximation. - Keep track of the decimal point when dividing inside the radical. - Use the thousandths digit to round to the nearest hundredth.

1. \(\frac{\sqrt{72}}{\sqrt{3}\sqrt{2}}=\sqrt{\frac{72}{6}}=\sqrt{12}=2\sqrt{3}\approx3.46\). 2. \(\sqrt{\frac{3}{4}}\sqrt{8}=\sqrt{6}\approx2.45\). 3. \(\frac{\sqrt{5.4}}{\sqrt{0.3}}=\sqrt{18}=3\sqrt{2}\approx4.24\). 4. \(\frac{\sqrt{14}\sqrt{2}}{\sqrt{4}}=\sqrt{\frac{28}{4}}=\sqrt{7}\approx2.65\).

1) \(2\sqrt{3}\approx3.46\) 2) \(\sqrt{6}\approx2.45\) 3) \(3\sqrt{2}\approx4.24\) 4) \(\sqrt{7}\approx2.65\)

- Can the quadratic expression under the root be written as a square plus a constant? - What is the smallest possible value of a square? - To check that every larger output occurs, can you work backward from a proposed output?

1. Complete the square: \(x^2-2x+5=(x-1)^2+4\). 2. Since \((x-1)^2\ge0\), the smallest value under the square root is \(4\), at \(x=1\). Thus \(f(x)\ge2\), with equality at \(x=1\). 3. Every \(y\ge2\) is attained: choose \(x=1+\sqrt{y^2-4}\), which is real, and then \(f(x)=\sqrt{(x-1)^2+4}=y\). 4. The range is \([2,\infty)\).

\([2,\infty)\). Since \(x^2-2x+5=(x-1)^2+4\ge4\), \(f(x)\ge2\), with equality at \(x=1\). For any \(y\ge2\), choosing \(x=1+\sqrt{y^2-4}\) gives \(f(x)=y\).

- Treat the powers of \(x\) and \(y\) separately. - Dividing like bases means subtracting exponents. - Move any factor with a negative exponent across the fraction bar at the end.

1. Subtract exponents on each common base. 2. For \(x\), \(-\frac{1}{2}-\frac{1}{4}=-\frac{3}{4}\). 3. For \(y\), \(\frac{2}{3}-(-\frac{1}{3})=1\). 4. The expression is \(x^{-\frac{3}{4}}y\), which with positive exponents is \(\frac{y}{x^{\frac{3}{4}}}\).

\(\frac{y}{x^{\frac{3}{4}}}\)

- Determine the sign of \(x^2\) before applying the fourth root. - Interpret \(x^{\frac{1}{2}}\) as a square root. - Compare the expressions only after identifying their domains.

1. Since \(x^2\ge0\) for every real \(x\), \((x^2)^{\frac{1}{4}}=\sqrt[4]{x^2}\) is defined for all real \(x\). 2. Also, \(\sqrt[4]{x^2}=\sqrt{|x|}\). 3. The expression \(x^{\frac{1}{2}}=\sqrt{x}\) is real only for \(x\ge0\). 4. On the common domain \(x\ge0\), \(|x|=x\), so the two expressions are equal. 5. For \(x<0\), only \((x^2)^{\frac{1}{4}}\) is real.

\((x^2)^{\frac{1}{4}}\) has domain \(\mathbb{R}\). \(x^{\frac{1}{2}}\) has domain \([0,\infty)\). They are equal for \(x\ge0\).

- In \(x^{m/n}\), decide which part controls the root index. - Use the negative test value in both expressions before comparing them. - An odd-index root and a square root have different real-domain behavior.

1. For \(x=-8\), \(x^{\frac{2}{3}}=(\sqrt[3]{-8})^2=(-2)^2=4\). 2. The claimed form gives \(\sqrt{(-8)^3}=\sqrt{-512}\), which is not real. 3. Mina reversed the roles of the numerator and denominator in the rational exponent. 4. A correct radical form is \(\sqrt[3]{x^2}\).

The claim is false. At \(x=-8\), \(x^{\frac{2}{3}}=4\), while \(\sqrt{x^3}\) is not real. The correct radical form is \(\sqrt[3]{x^2}\).

- Evaluate each side separately for the given values. - Expand \((x+y)^2\). - Compare the squared expressions and use the fact that both original sides are positive.

1. For a), \(\sqrt{9}+\sqrt{16}=3+4=7\), while \(\sqrt{9+16}=\sqrt{25}=5\). The equation is false for these values. 2. For positive \(a\) and \(b\), \((\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{ab}\). 3. Also, \((\sqrt{a+b})^2=a+b\). 4. Since \(a>0\) and \(b>0\), \(2\sqrt{ab}>0\). Therefore, \((\sqrt{a}+\sqrt{b})^2>(\sqrt{a+b})^2\). Both original expressions are positive, so \(\sqrt{a}+\sqrt{b}>\sqrt{a+b}\). 5. Thus, Arjun's conjecture is correct for all positive \(a\) and \(b\).

a) The equation is false because \(7\ne5\). b) The conjecture is correct. For \(a,b>0\), squaring gives \(a+b+2\sqrt{ab}>a+b\), so \(\sqrt{a}+\sqrt{b}>\sqrt{a+b}\).

- A real square root requires a nonnegative radicand. - A denominator cannot equal zero. - A minus sign before a variable does not by itself determine the sign of its value. - Determine the signs of the product or quotient after imposing the domain restrictions.

1. For \(\sqrt{-a}\) and \(\sqrt{-b}\) to be real, \(a\le0\) and \(b\le0\). Then \(ab\ge0\), so the left side is also defined. Under these conditions, \(\sqrt{-a}\cdot\sqrt{-b}=\sqrt{(-a)(-b)}=\sqrt{ab}\). 2. The numerator requires \(a\le0\). The denominator requires \(-b>0\), so \(b<0\). Then \(\frac{a}{b}\ge0\), and \(\frac{\sqrt{-a}}{\sqrt{-b}}=\sqrt{\frac{-a}{-b}}=\sqrt{\frac{a}{b}}\).

1) \(a\le0\) and \(b\le0\) 2) \(a\le0\) and \(b<0\)

- Combine the fractions and factor the numerator under the radical. - Enforce both the nonnegative-radicand condition and the nonzero-denominator condition. - Check separately whether a factor that is usually positive can equal zero. - Extract the fourth powers from the square root.

1. Combine and factor the terms under the radical: \(T=\sqrt{\frac{a^5-a^4}{b^4}}=\sqrt{\frac{a^4(a-1)}{b^4}}\). 2. The denominator requires \(b\ne0\). Because \(a^4\ge0\), the radicand is nonnegative when \(a=0\) or \(a-1\ge0\). Thus, \(a\in\{0\}\cup[1,\infty)\). 3. For \(a\ge1\), \(T=\sqrt{\frac{a^4}{b^4}}\sqrt{a-1}=\frac{a^2}{b^2}\sqrt{a-1}\). 4. When \(a=0\), the original radicand is \(0\), so \(T=0\).

a) For \(a\ge1\), \(T=\frac{a^2}{b^2}\sqrt{a-1}\); for \(a=0\), \(T=0\). b) \(a\in\{0\}\cup[1,\infty)\) and \(b\ne0\)

- Which exponent would turn the denominator into a perfect \(n\)th power? - How can you multiply the radicand by \(1\) without changing it? - What happens if the denominator already contains a perfect \(n\)th power?

1. Multiply inside the radical by \(\frac{a^{n-k}}{a^{n-k}}\), giving \(T=\frac{\sqrt[n]{xa^{n-k}}}{a}\). 2. With \(n=5\) and \(k=2\), obtain \(\frac{\sqrt[5]{2a^3}}{a}\). 3. If \(k=mn\), then \(\sqrt[n]{\frac{x}{a^{mn}}}=\frac{\sqrt[n]{x}}{a^m}\). For example, \(\sqrt[3]{\frac{5}{2^6}}=\frac{\sqrt[3]{5}}{4}\).

a) \(\frac{\sqrt[n]{xa^{n-k}}}{a}\). b) \(\frac{\sqrt[5]{2a^3}}{a}\). c) Yes. For example, \(\sqrt[3]{\frac{5}{2^6}}=\frac{\sqrt[3]{5}}{4}\).

- Move the entire outside fraction inside the fourth root. - Simplify all numerical and variable powers in the radicand. - Multiply the radicand fraction so its denominator becomes a perfect fourth power.

1. Move the outside factor inside the fourth root: \(T=\sqrt[4]{\left(\frac{2a^2}{3b}\right)^4\frac{27b^5}{8a^6}}\). 2. Simplify the radicand: \(\left(\frac{2a^2}{3b}\right)^4\frac{27b^5}{8a^6}=\frac{16a^8}{81b^4}\cdot\frac{27b^5}{8a^6}=\frac{2a^2b}{3}\). 3. Make the denominator a fourth power: \(\sqrt[4]{\frac{2a^2b}{3}}=\sqrt[4]{\frac{54a^2b}{81}}=\frac13\sqrt[4]{54a^2b}\).

\(\frac13\sqrt[4]{54a^2b}\)

- Which factors are perfect \(k\)th powers for every allowed \(k\)? - After canceling outside factors, consider the small allowed values of \(k\). - When can the remaining radical be simplified further?

1. Rewrite the radicand as \(x^kx^3(y^2)^k\), then extract \(xy^2\). 2. The expression becomes \(2\sqrt[k]{x^3}\). 3. For \(k=2\), simplify further to \(2x\sqrt{x}\). For \(k=3\), simplify further to \(2x\). For \(k\ge4\), \(2\sqrt[k]{x^3}\) is fully simplified.

For \(k=2\): \(2x\sqrt{x}\). For \(k=3\): \(2x\). For \(k\ge4\): \(2\sqrt[k]{x^3}\).

- Simplify the fraction outside the radical first. - Use the difference-of-squares identity. - Rewrite the radical denominator as a rational exponent and subtract exponents.

1. Use the difference of squares: \(\frac{a^2-b^2}{a+b}=a-b\). 2. Because \(a>b>0\), \(\sqrt[n]{\frac{(a+b)^n}{(a-b)^{n-1}}}=\frac{a+b}{(a-b)^{\frac{n-1}{n}}}\). 3. Multiply and combine powers: \((a-b)\frac{a+b}{(a-b)^{\frac{n-1}{n}}}=(a+b)(a-b)^{\frac1n}=(a+b)\sqrt[n]{a-b}\).

\((a+b)\sqrt[n]{a-b}\)

- Extract powers whose exponents are multiples of the radical index. - For a fraction inside a radical, make the denominator a perfect power. - Like radical terms have the same index and the same simplified radicand.

1. The first expression is \(a\sqrt[n]{a^2b}\). The second already has that form. The third is \(\frac1a\cdot a^2\sqrt[n]{a^2b}=a\sqrt[n]{a^2b}\). All three are identical. 2. \(\sqrt[4]{\frac{x^5}{y^3}}=\frac{x}{y}\sqrt[4]{xy}\). The second expression already has that radical part. The third is \(\frac1x\cdot x^2\sqrt[4]{xy}=x\sqrt[4]{xy}\). All three have the same radical part \(\sqrt[4]{xy}\), so they are like radical terms.

1) All three terms are identical and therefore like terms. 2) All three terms are like radical terms with radical part \(\sqrt[4]{xy}\).

- What powers can come outside the first radical? - Would fractional exponents help you rewrite the second radical with matching powers? - Compare the remaining radical factors, not the outside coefficients.

1. Extract the perfect power from \(A\): \(A=x^2\sqrt[k]{xy}\). 2. Rewrite \(B\) using fractional powers: \(B=x^{\frac{1}{k}}y^{-\frac{k-1}{k}}=\frac{1}{y}x^{\frac{1}{k}}y^{\frac{1}{k}}=\frac{1}{y}\sqrt[k]{xy}\). 3. Both terms have the radical part \(\sqrt[k]{xy}\), so Mateo is correct.

\(A=x^2\sqrt[k]{xy}\) and \(B=\frac{1}{y}\sqrt[k]{xy}\). Mateo is correct: both have the radical part \(\sqrt[k]{xy}\).

- Substitute a new variable for the square root. - Treat the resulting expression as a quadratic polynomial. - Look for a perfect-square trinomial in part 1. - Substitute the original radical expression back after factoring.

1. Let \(u=\sqrt{x}\). Then the expression is \(u^2-10u+25=(u-5)^2\). Substituting back gives \((\sqrt{x}-5)^2\). 2. Let \(u=\sqrt{a}\). Then \(3a-5\sqrt{a}-2=3u^2-5u-2\). Factor the quadratic: \(3u^2-5u-2=(u-2)(3u+1)\). Substituting back gives \((\sqrt{a}-2)(3\sqrt{a}+1)\).

1) \((\sqrt{x}-5)^2\) 2) \((\sqrt{a}-2)(3\sqrt{a}+1)\)

- Square the simpler radical expression. - Compare the rational part and the radical part after squaring. - Use the sum and product of \(x\) and \(y\). - Compare coefficients carefully when radical terms have different outside factors.

1. \((\sqrt{7}-\sqrt{3})^2=7-2\sqrt{21}+3=10-\sqrt{84}\), and \(\sqrt{7}-\sqrt{3}>0\). Therefore, the equation is true. 2. Squaring the proposed form gives \(x+y+2\sqrt{xy}=4.5+2\sqrt{2}\). Thus, \(x+y=4.5\) and \(xy=2\). The values are \(x=4\) and \(y=0.5\), in either order. 3. For nonnegative \(x\) and \(y\), \((\sqrt{x}+\sqrt{y})^2=x+y+2\sqrt{xy}\). Therefore, the required conditions are \(x+y=a\) and \(xy=b\).

1. The equation is true. 2. \(x=4\) and \(y=0.5\), in either order 3. \(x,y\ge0\), \(x+y=a\), and \(xy=b\)

- Find \(x-y\) and \(x+y\) separately using the same denominator. - Expand the squared binomials carefully. - Which factors appear in both the numerator and denominator of the final complex fraction?

1. Using common denominator \(a-1\), \(x-y=\frac{(\sqrt{a}+1)^2-(\sqrt{a}-1)^2}{a-1}=\frac{4\sqrt{a}}{a-1}\). 2. Similarly, \(x+y=\frac{(\sqrt{a}+1)^2+(\sqrt{a}-1)^2}{a-1}=\frac{2(a+1)}{a-1}\). 3. Divide to obtain \(S=\frac{2\sqrt{a}}{a+1}\).

\(x-y=\frac{4\sqrt{a}}{a-1}\) and \(x+y=\frac{2(a+1)}{a-1}\); dividing gives \(S=\frac{2\sqrt{a}}{a+1}\).

- Substitute the expression for \(x\) into \(x^2-1\) and rewrite the result as a perfect square. - Use \(a>1\) to determine the sign when taking the square root. - Simplify the numerator and denominator of \(E\) separately.

1. Substitute for \(x\): \(x^2-1=\frac{1}{4}\left(a+2+\frac{1}{a}\right)-1=\frac{1}{4}\left(a-2+\frac{1}{a}\right)=\left[\frac{1}{2}\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\right]^2\). 2. Since \(a>1\), \(\sqrt{a}-\frac{1}{\sqrt{a}}>0\). Thus, \(\sqrt{x^2-1}=\frac{1}{2}\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\). 3. Then \(x+\sqrt{x^2-1}=\sqrt{a}\), while \(x-\sqrt{x^2-1}=\frac{1}{\sqrt{a}}\). 4. Therefore, \(E=\frac{\sqrt{a}}{1/\sqrt{a}}=a\).

\(E=a\)

- Isolate the radical in the equation defining \(x\), then square. - Use the resulting quadratic relation to reduce higher powers of \(x\). - Rewrite the cubic expression to include a factor known to equal zero. - Consider polynomial division or strategic factoring if a cubic remains.

1. From \(2x-1=\sqrt{3}\), squaring gives \(4x^2-4x+1=3\), so \(2x^2-2x-1=0\). 2. Rewrite the expression as \(x(2x^2-2x-1)+4x+2\). 3. The factor in parentheses equals \(0\), so the value is \(4x+2\). 4. Substituting gives \(4\left(\frac{1+\sqrt{3}}{2}\right)+2=4+2\sqrt{3}\).

\(4+2\sqrt{3}\)

- Rewrite \(x+2\) and \(x-2\) over the common denominator \(ab\). - Recognize perfect-square trinomials in the numerators. - Use \(a>b>0\) when taking square roots. - Simplify the numerator and denominator of the complex fraction separately.

1. Substitute for \(x\): \(x+2=\frac{a^2+b^2+2ab}{ab}=\frac{(a+b)^2}{ab}\), and \(x-2=\frac{a^2+b^2-2ab}{ab}=\frac{(a-b)^2}{ab}\). 2. Since \(a>b>0\), \(\sqrt{x+2}=\frac{a+b}{\sqrt{ab}}\) and \(\sqrt{x-2}=\frac{a-b}{\sqrt{ab}}\). 3. The numerator of \(T\) is \(\frac{2a}{\sqrt{ab}}\), and the denominator is \(\frac{2b}{\sqrt{ab}}\). 4. Therefore, \(T=\frac{2a/\sqrt{ab}}{2b/\sqrt{ab}}=\frac{a}{b}\).

\(\frac{a}{b}\)

- Use \(0<x<y\) to determine the sign of the outside factor. - Keep a negative sign outside when moving the positive magnitude under the radical. - Factor the difference of squares in the denominator. - Cancel common factors inside the radicand.

1. Since \(0<x<y\), \(\frac{x-y}{x}<0\). Rewrite it as \(-\frac{y-x}{x}\), where \(\frac{y-x}{x}>0\). 2. Move the positive factor under the radical by squaring it: \(-\sqrt{\frac{(y-x)^2}{x^2}\cdot\frac{x^3}{y^2-x^2}}\). 3. Factor \(y^2-x^2=(y-x)(y+x)\), then cancel common factors. The radicand becomes \(\frac{x(y-x)}{x+y}\). 4. Therefore, the simplified expression is \(-\sqrt{\frac{x(y-x)}{x+y}}\).

\(-\sqrt{\frac{x(y-x)}{x+y}}\)