- Choose a factor that turns the denominator into a perfect square under the radical. - The same factor must multiply numerator and denominator.

1. The denominator contains \(\sqrt{6}\). 2. Multiplying by another \(\sqrt{6}\) makes the denominator \(6\), which is rational.

\(\sqrt{6}\)

- Keep both terms unchanged. - Change only the operation between the terms.

1. A conjugate keeps the same two terms and changes the sign between them. 2. Therefore, the conjugate of \(3+\sqrt{5}\) is \(3-\sqrt{5}\).

\(3-\sqrt{5}\)

- Factor the radicand into prime powers. - For a cube root, each exponent in the denominator radicand should become a multiple of \(3\).

1. Factor the radicand: \(12=2^2\cdot3\). 2. To make a perfect cube, one more factor of \(2\) and two more factors of \(3\) are needed. 3. The missing factor is \(2\cdot3^2=18\), so the needed radical factor is \(\sqrt[3]{18}\).

\(\sqrt[3]{18}\)

- Focus on what is below the fraction bar, not what is above it. - Which of the denominators is not a rational number?

1. The denominator in a) is \(\sqrt{5}\), which is irrational. 2. The denominators in b) and c) are the rational numbers \(5\) and \(3\).

a) only.

- What happens to the two middle terms when you expand these binomials? - How does squaring a square root affect the result?

1. Multiply using difference of squares: \((3+\sqrt{2})(3-\sqrt{2})=3^2-(\sqrt{2})^2\). 2. This equals \(9-2=7\), a whole number.

\(7\). The radical terms cancel in the conjugate product.

- When does the real square root of a number exist? - Can a denominator used in a factor of one be zero?

1. For a real denominator, \(x\ge0\) is needed. 2. At \(x=0\), the original denominator and the proposed multiplier's denominator are zero. 3. Thus \(x>0\) makes the multiplier a valid factor of \(1\).

At \(x=0\), the denominator is zero. Only \(x>0\) gives a real, nonzero \(\sqrt{x}\).

- Which choice makes a difference-of-squares product? - Think about whether any radical terms will remain after multiplying.

1. Multiplying by \(\sqrt{3}\) gives \(2\sqrt{3}+3\), which is irrational. 2. The conjugate gives \((2+\sqrt{3})(2-\sqrt{3})=4-3=1\), which is rational.

Multiply by \(2-\sqrt{3}\): \((2+\sqrt{3})(2-\sqrt{3})=1\). The other choice leaves \(2\sqrt{3}\) in the product.

- The denominator is a difference of two square roots. - Choose a factor that turns the denominator into a difference of squares. - Multiply the numerator and denominator by the same expression.

1. Use the conjugate \(\sqrt{7}+\sqrt{5}\) as the rationalizing factor. 2. Multiply by \(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}\). 3. The denominator becomes \((\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})=7-5=2\). 4. The numerator becomes \(\sqrt{7}(\sqrt{7}+\sqrt{5})=7+\sqrt{35}\). 5. Therefore, the rationalized expression is \(\frac{7+\sqrt{35}}{2}\).

Rationalizing factor: \(\sqrt{7}+\sqrt{5}\) \(\frac{\sqrt{7}}{\sqrt{7}-\sqrt{5}}=\frac{7+\sqrt{35}}{2}\).

- Multiply each fraction by a form of \(1\) that removes the radical from its denominator. - Record the rationalized exact form of each quantity before comparing them. - Once both quantities are positive, a comparison of their squares preserves their order.

1. Rationalize \(A\): \(A=\frac{12\sqrt{6}}{6}=2\sqrt{6}\). 2. Rationalize \(B\): \(B=\frac{15\sqrt{5}}{5}=3\sqrt{5}\). 3. Both values are positive. Their squares are \(A^2=24\) and \(B^2=45\). 4. Since \(45>24\), \(B>A\).

\(A=2\sqrt{6}\) and \(B=3\sqrt{5}\). Therefore, \(B>A\).

- Multiply by a radical that makes the denominator a perfect square. - Simplify radicals and common factors before or after rationalizing. - When the numerator has more than one term, multiply every term by the radical factor.

1. In a), \(\frac{7}{\sqrt{7}}\cdot\frac{\sqrt{7}}{\sqrt{7}}=\frac{7\sqrt{7}}{7}=\sqrt{7}\). 2. In b), first simplify \(\frac{\sqrt{3}}{2\sqrt{6}}=\frac{1}{2\sqrt{2}}\). Then rationalize: \(\frac{1}{2\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{4}\). 3. In c), \(\frac{4-\sqrt{2}}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{4\sqrt{2}-2}{2}=2\sqrt{2}-1\). 4. In d), \(\sqrt{75}=5\sqrt{3}\), so \(\frac{15}{\sqrt{75}}=\frac{3}{\sqrt{3}}=\sqrt{3}\).

a) \(\sqrt{7}\) b) \(\frac{\sqrt{2}}{4}\) c) \(2\sqrt{2}-1\) d) \(\sqrt{3}\)

- Simplify the radical in the denominator before rationalizing when possible. - Multiply the numerator and denominator by the denominator radical. - Distribute the radical across every term in the numerator. - Check for common factors after rationalizing.

1. In a), \(\frac{10}{\sqrt{20}}=\frac{10}{2\sqrt{5}}=\frac{5}{\sqrt{5}}=\sqrt{5}\). 2. In b), multiply by \(\frac{\sqrt{5}}{\sqrt{5}}\): \(\frac{(\sqrt{5}+2)\sqrt{5}}{15}=\frac{5+2\sqrt{5}}{15}\).

a) \(\sqrt{5}\) b) \(\frac{5+2\sqrt{5}}{15}\)

- Multiply by the conjugate of the denominator. - Apply the difference-of-squares identity. - Keep common factors visible before canceling.

1. Multiply the numerator and denominator by the conjugate \(4+\sqrt{5}\). 2. The denominator becomes \((4-\sqrt{5})(4+\sqrt{5})=16-5=11\). 3. Therefore, \(\frac{11}{4-\sqrt{5}}=\frac{11(4+\sqrt{5})}{11}=4+\sqrt{5}\).

\(4+\sqrt{5}\)

- Rationalize a denominator by multiplying by a suitable radical form of \(1\). - Extract perfect-square factors from radicals. - Use the quotient property when both numerator and denominator are square roots. - Simplify common factors at the end.

1. In a), \(\frac{14}{\sqrt{7}}\cdot\frac{\sqrt{7}}{\sqrt{7}}=\frac{14\sqrt{7}}{7}=2\sqrt{7}\). 2. In b), \(\sqrt{32}=4\sqrt{2}\), so \(\frac{\sqrt{32}}{4}=\sqrt{2}\). 3. In c), \(\sqrt{20}=2\sqrt{5}\), so \(\frac{1}{\sqrt{20}}=\frac{1}{2\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{5}}{10}\). 4. In d), \(\frac{\sqrt{75}}{\sqrt{3}}=\sqrt{\frac{75}{3}}=\sqrt{25}=5\).

a) \(2\sqrt{7}\) b) \(\sqrt{2}\) c) \(\frac{\sqrt{5}}{10}\) d) \(5\)

- Multiply by a radical form of \(1\) that makes the denominator a square. - Multiply both the numerator and denominator by the same radical expression. - Simplify the product of a square root with itself. - Cancel common factors only after confirming they are nonzero.

1. In a), \(\frac{12}{\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}}=2\sqrt{6}\). 2. In b), \(\frac{5}{2\sqrt{10}}\cdot\frac{\sqrt{10}}{\sqrt{10}}=\frac{5\sqrt{10}}{20}=\frac{\sqrt{10}}{4}\). 3. In c), \(\frac{z^2}{\sqrt{z}}\cdot\frac{\sqrt{z}}{\sqrt{z}}=\frac{z^2\sqrt{z}}{z}=z\sqrt{z}\). 4. In d), \(\frac{6k}{\sqrt{2k}}\cdot\frac{\sqrt{2k}}{\sqrt{2k}}=\frac{6k\sqrt{2k}}{2k}=3\sqrt{2k}\).

a) \(2\sqrt{6}\) b) \(\frac{\sqrt{10}}{4}\) c) \(z\sqrt{z}\) d) \(3\sqrt{2k}\)

- Multiply by a radical that completes the denominator exponent to the radical index. - Use the product property for radicals with the same index. - Cancel any common numerical factor after rationalizing.

1. Multiply by \(\frac{\sqrt[3]{2^2}}{\sqrt[3]{2^2}}\): \(\frac{4\sqrt[3]{4}}{2}=2\sqrt[3]{4}\). 2. Multiply by \(\frac{\sqrt[4]{5^3}}{\sqrt[4]{5^3}}\): \(\frac{\sqrt[4]{125}}{5}\). 3. Multiply by \(\frac{\sqrt[5]{3^3}}{\sqrt[5]{3^3}}\): \(\frac{3\sqrt[5]{27}}{3}=\sqrt[5]{27}\).

1) \(2\sqrt[3]{4}\) 2) \(\frac{\sqrt[4]{125}}{5}\) 3) \(\sqrt[5]{27}\)

- Use the conjugate of a binomial radical denominator. - Apply the difference-of-squares identity. - In b), rewrite \(x\) as \((\sqrt{x})^2\) before factoring.

1. In a), multiply by the conjugate \(\sqrt{7}-\sqrt{2}\): \(\frac{10(\sqrt{7}-\sqrt{2})}{7-2}=2(\sqrt{7}-\sqrt{2})\). 2. In b), factor the numerator as a difference of squares: \(x-25=(\sqrt{x}-5)(\sqrt{x}+5)\). 3. Since \(x\ne25\), the common factor \(\sqrt{x}-5\) is nonzero and can be canceled. The result is \(\sqrt{x}+5\).

a) \(2(\sqrt{7}-\sqrt{2})\) b) \(\sqrt{x}+5\)

- Use the conjugate of a denominator containing a radical difference. - Apply the difference-of-squares identity. - In b), rewrite \(a\) as \((\sqrt{a})^2\). - Check that a factor is nonzero before canceling it.

1. In a), multiply by the conjugate \(\sqrt{x+1}+1\): \(\frac{x(\sqrt{x+1}+1)}{(x+1)-1}=\frac{x(\sqrt{x+1}+1)}{x}\). Since \(x\ne0\), canceling gives \(\sqrt{x+1}+1\). 2. In b), factor the numerator: \(a-16=(\sqrt{a}-4)(\sqrt{a}+4)\). Since \(\sqrt{a}+4>0\), canceling gives \(\sqrt{a}-4\).

a) \(\sqrt{x+1}+1\) b) \(\sqrt{a}-4\)

- Use the conjugate of the denominator. - Apply the difference-of-squares identity. - Use the condition \(x\ne y\) to confirm that the new denominator is nonzero.

1. Multiply by the conjugate \(\sqrt{x}-\sqrt{y}\). 2. The denominator becomes \((\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})=x-y\). 3. Therefore, the rationalized form is \(\frac{\sqrt{x}-\sqrt{y}}{x-y}\).

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

- Use the conjugate of each binomial denominator. - Apply the difference-of-squares identity. - Multiply the numerator and denominator by the same expression so the fraction’s value is unchanged. - Cancel numerical factors before expanding the numerator when possible.

1. Multiply by the conjugate \(\sqrt{10}+\sqrt{7}\): \(\frac{6(\sqrt{10}+\sqrt{7})}{10-7}=2\sqrt{10}+2\sqrt{7}\). 2. Multiply by \(\sqrt{5}-\sqrt{3}\): \(\frac{2\sqrt{5}(\sqrt{5}-\sqrt{3})}{5-3}=5-\sqrt{15}\).

1) \(2\sqrt{10}+2\sqrt{7}\) 2) \(5-\sqrt{15}\)

- Multiply by the conjugate of the denominator. - Apply the difference-of-squares identity in the denominator. - Distribute carefully when multiplying the numerator binomials.

1. Multiply by the conjugate \(2\sqrt{3}+1\). 2. The denominator is \((2\sqrt{3})^2-1^2=12-1=11\). 3. The numerator is \((\sqrt{3}+2)(2\sqrt{3}+1)=8+5\sqrt{3}\). 4. Therefore, the result is \(\frac{8+5\sqrt{3}}{11}\).

\(\frac{8+5\sqrt{3}}{11}\)

- Multiply each binomial denominator by its conjugate. - Apply the difference-of-squares identity. - Cancel the resulting rational denominator with the numerator factor when possible.

1. In a), multiply by \(4+\sqrt{5}\): \(\frac{22(4+\sqrt{5})}{16-5}=2(4+\sqrt{5})=8+2\sqrt{5}\). 2. In b), multiply by \(2\sqrt{6}-3\): \(\frac{15(2\sqrt{6}-3)}{24-9}=2\sqrt{6}-3\).

a) \(8+2\sqrt{5}\) b) \(2\sqrt{6}-3\)

- Rationalize the fraction using the conjugate of its denominator. - Simplify the rationalized fraction before subtracting \(\sqrt{7}\). - Identify the number set containing the final value.

1. Rationalize the denominator: \(\frac{6}{\sqrt{7}-1}\cdot\frac{\sqrt{7}+1}{\sqrt{7}+1}=\frac{6(\sqrt{7}+1)}{7-1}=\sqrt{7}+1\). 2. Therefore, \((\sqrt{7}+1)-\sqrt{7}=1\), which is an integer.

\(1\)

- Which factor would turn the radical in the denominator into a whole number? - Whatever factor you choose, multiply both numerator and denominator by it.

1. Multiply numerator and denominator by \(\sqrt{3}\) to obtain \(\frac{5\sqrt{3}}{2(\sqrt{3})^2}\). 2. Since \((\sqrt{3})^2=3\), the result is \(\frac{5\sqrt{3}}{6}\).

\(\frac{5\sqrt{3}}{6}\)

- What kind of two-term product removes the square roots from the denominator? - The numerator must be multiplied by the same nonzero expression as the denominator.

1. Multiply by the conjugate over itself to get \(\frac{\sqrt{5}+\sqrt{3}}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}\). 2. The denominator is \(5-3=2\), so the result is \(\frac{\sqrt{5}+\sqrt{3}}{2}\).

\(\frac{\sqrt{5}+\sqrt{3}}{2}\)

- Which inputs make the square root in the denominator real and nonzero? - What can you multiply the denominator by to remove its square root? - Use the same input in both forms to check the equality.

1. The radical in the denominator is real and nonzero when \(x+1>0\), or \(x>-1\). 2. Multiply numerator and denominator by \(\sqrt{x+1}\): \(\frac{(x+2)\sqrt{x+1}}{x+1}\). 3. At \(x=3\), both the original expression and the rationalized one equal \(\frac{5}{2}\).

\(\frac{(x+2)\sqrt{x+1}}{x+1}\), with \(x>-1\); at \(x=3\), both forms give \(\frac{5}{2}\).

- Use area \(=(\text{side})^2\) for the first square. - Simplify \(\sqrt{18}\) before rationalizing the ratio. - For a denominator \(\sqrt{2}\), choose a factor that makes its product a rational number.

1. \(A_1=a_1^2=(\sqrt{18}\,\text{cm})^2=18\,\text{cm}^2\). 2. \(A_2=2A_1=36\,\text{cm}^2\), so \(a_2=\sqrt{36}\,\text{cm}=6\,\text{cm}\). 3. \(\frac{a_2}{a_1}=\frac{6}{\sqrt{18}}=\frac{2}{\sqrt{2}}\). 4. Multiply numerator and denominator by \(\sqrt{2}\): \(\frac{2}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{2\sqrt{2}}{2}=\sqrt{2}\).

1. \(A_1=18\,\text{cm}^2\) 2. \(a_2=6\,\text{cm}\) 3. The rationalizing factor is \(\sqrt{2}\), and \(\frac{a_2}{a_1}=\sqrt{2}\).

- Use the area formula to solve for the unknown leg. - What can you multiply a radical denominator by to make it rational? - Use the given approximation only after simplifying the exact expression.

1. The area formula gives \(\frac{6}{\sqrt{2}}=\frac{1}{2}\cdot3\cdot b\), so \(b=\frac{4}{\sqrt{2}}\,\text{cm}\). 2. Multiply the numerator and denominator by \(\sqrt{2}\): \(b=2\sqrt{2}\,\text{cm}\). 3. The approximation is \(b\approx2.82\,\text{cm}\).

From \(\frac{3b}{2}=\frac{6}{\sqrt{2}}\), we get \(b=\frac{4}{\sqrt{2}}=2\sqrt{2}\,\text{cm}\approx2.82\,\text{cm}\).

- Multiply the numerator and denominator by the conjugate of \(\sqrt{11}-3\). - Use the difference-of-squares identity in the denominator. - Use the thousandths digit to round to the nearest hundredth.

1. Multiply by the conjugate: \(T=\frac{4}{\sqrt{11}-3}\cdot\frac{\sqrt{11}+3}{\sqrt{11}+3}\). 2. The denominator is \((\sqrt{11})^2-3^2=11-9=2\), so \(T=\frac{4(\sqrt{11}+3)}{2}=2(\sqrt{11}+3)\). 3. Using the given approximation, \(T\approx2(3.317+3)=12.634\), which rounds to \(12.63\).

a) \(2(\sqrt{11}+3)\) b) \(12.63\)

- Remove the radical from the denominator of \(x\) before making any comparison. - Look for a perfect-square factor in \(192\). - For the remaining positive comparison, squaring preserves order.

1. Rationalize \(x\): \(x=\frac{18\sqrt{3}}{3}=6\sqrt{3}\). 2. Simplify \(z\): \(z=\sqrt{64\cdot3}=8\sqrt{3}\). 3. Thus, \(x<z\) because \(6\sqrt{3}<8\sqrt{3}\). 4. Compare \(z\) and \(y\), which are positive: \(z^2=192\) and \(y^2=200\), so \(z<y\). 5. Therefore, \(x<z<y\).

\(x=6\sqrt{3}\) and \(z=8\sqrt{3}\). The order is \(\frac{18}{\sqrt{3}}<\sqrt{192}<10\sqrt{2}\).

- Identify which denominator contains a radical in each part. - Choose a factor that makes that denominator rational when multiplied by itself. - After the denominator is rational, simplify any remaining radicals and combine like terms.

1. In a), rationalize \(\frac{3}{\sqrt{3}}\) by multiplying by \(\frac{\sqrt{3}}{\sqrt{3}}\): \(\frac{3\sqrt{3}}{3}=\sqrt{3}\). Also, \(\sqrt{12}=2\sqrt{3}\). Therefore, the result is \(\sqrt{3}\). 2. In b), multiply numerator and denominator by \(\sqrt{5}\): \(\frac{(\sqrt{10}+\sqrt{20})\sqrt{5}}{5}=\frac{\sqrt{50}+\sqrt{100}}{5}=\frac{5\sqrt{2}+10}{5}=\sqrt{2}+2\). 3. In c), multiply by \(\frac{\sqrt{2}}{\sqrt{2}}\): \(\frac{2\sqrt{2}}{2}=\sqrt{2}\).

a) Rationalizing factor: \(\frac{\sqrt{3}}{\sqrt{3}}\); result: \(\sqrt{3}\) b) Rationalizing factor: \(\frac{\sqrt{5}}{\sqrt{5}}\); result: \(2+\sqrt{2}\) c) Rationalizing factor: \(\frac{\sqrt{2}}{\sqrt{2}}\); \(\frac{2}{\sqrt{2}}=\sqrt{2}\)

- Which multiplier makes a binomial radical denominator rational? - Compare the squared value in part b) with neighboring perfect squares. - Use the rectangle area formula to find the missing side.

1. For a), multiplying by \(\frac{2-\sqrt{3}}{2-\sqrt{3}}\) gives \(2-\sqrt{3}\). 2. For b), \(\frac{12}{\sqrt{3}}=4\sqrt{3}=\sqrt{48}\), which is between \(6\) and \(7\). 3. For c), length is \(\frac{10}{\sqrt{5}}=2\sqrt{5}\,\text{cm}\).

a) \(\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}=2-\sqrt{3}\). b) \(\frac{12}{\sqrt{3}}=4\sqrt{3}=\sqrt{48}\), between \(6\) and \(7\). c) \(\frac{10}{\sqrt{5}}=\frac{10\sqrt{5}}{5}=2\sqrt{5}\,\text{cm}\).

- Work on the fraction before combining it with the final radical term. - Which factor would remove the radical from the denominator? - Use a binomial product formula in both numerator and denominator.

1. Multiply the fraction by \(\frac{\sqrt{3}-1}{\sqrt{3}-1}\). 2. The numerator becomes \(4-2\sqrt{3}\), and the denominator becomes \(2\), so the fraction equals \(2-\sqrt{3}\). 3. Therefore, \(T=2\).

\(\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)^2}{3-1}=2-\sqrt{3}\); adding \(\sqrt{3}\) gives \(T=2\).

- Rationalize the first denominator using its conjugate. - Use the difference-of-squares identity in the conjugate product. - Extract the largest perfect-square factor from \(48\). - Combine like radical terms after simplifying each part.

1. Rationalize the denominator: \(\frac{4}{1-\sqrt{3}}\cdot\frac{1+\sqrt{3}}{1+\sqrt{3}}=\frac{4(1+\sqrt{3})}{1-3}=-2-2\sqrt{3}\). 2. Simplify \(\sqrt{48}=4\sqrt{3}\). 3. Combine like radical terms: \((-2-2\sqrt{3})+4\sqrt{3}+2\sqrt{3}=-2+4\sqrt{3}\).

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

- Factor a perfect square from each radicand. - Convert the mixed number to an improper fraction. - Multiply by a suitable radical to remove the radical from the denominator. - Simplify common factors in the quotient.

1. \(A=\sqrt{180}=\sqrt{36\cdot5}=6\sqrt{5}\). 2. Since \(3\frac{1}{5}=\frac{16}{5}\), \(B=\sqrt{\frac{16}{5}}=\frac{4}{\sqrt{5}}=\frac{4\sqrt{5}}{5}\). 3. Both simplified expressions are rational multiples of \(\sqrt{5}\), so they are like radicals. 4. \(\frac{A}{B}=\frac{6\sqrt{5}}{4\sqrt{5}/5}=\frac{30}{4}=\frac{15}{2}=7.5\).

a) \(A=6\sqrt{5}\) and \(B=\frac{4\sqrt{5}}{5}\) b) Yes. Both are multiples of \(\sqrt{5}\). c) \(\frac{A}{B}=\frac{15}{2}=7.5\)

- Use a conjugate when the denominator is a sum or difference of radicals. - Apply the difference-of-squares identity in the denominator. - Multiply the entire numerator and denominator by the conjugate. - Simplify quotients of radicals and any common numerical factors.

1. In a), multiply by the conjugate \(\sqrt{5}+\sqrt{2}\): \(\frac{6(\sqrt{5}+\sqrt{2})}{5-2}=2\sqrt{5}+2\sqrt{2}\). 2. In b), multiply by \(\sqrt{3}+1\): \(\frac{(\sqrt{3}+1)^2}{3-1}=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}\). 3. In c), \(\frac{3\sqrt{2}}{\sqrt{6}}=3\sqrt{\frac{2}{6}}=\frac{3}{\sqrt{3}}=\sqrt{3}\).

a) \(2\sqrt{5}+2\sqrt{2}\) b) \(2+\sqrt{3}\) c) \(\sqrt{3}\)

- Which multiplier turns the denominator into a rational number? - Compare division by a decimal with multiplication by a whole number. - Round only after carrying out each computation.

1. Direct division gives \(10\div1.414\approx7.0721\), which rounds to \(7.07\). 2. Rationalization gives \(T=5\sqrt{2}\). Then \(5\cdot1.414=7.07\). 3. Both methods agree to two decimal places, and multiplication by \(5\) is simpler without a calculator.

1) \(10\div1.414\approx7.07\). 2) \(\frac{10}{\sqrt{2}}=5\sqrt{2}\approx5\cdot1.414=7.07\). Rationalizing first is easier by hand because it uses multiplication instead of decimal division.

- Multiply by the conjugate of the denominator. - Apply the difference-of-squares identity. - Substitute the decimal approximations only after simplifying exactly.

1. Multiply by the conjugate: \(E=\frac{2}{\sqrt{5}-\sqrt{3}}\cdot\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}\). 2. The denominator is \(5-3=2\), so \(E=\sqrt{5}+\sqrt{3}\). 3. Using the given approximations, \(E\approx2.236+1.732=3.968\), which rounds to \(3.97\).

a) \(\sqrt{5}+\sqrt{3}\) b) \(3.97\)

- Complete the exponent in each radical denominator to the radical index. - Multiply numerator and denominator by the same radical. - Cancel any common factor after rationalizing.

1. Multiply by \(\frac{\sqrt[n]{v^{n-1}}}{\sqrt[n]{v^{n-1}}}\): \(\frac{u\sqrt[n]{v^{n-1}}}{v}\). 2. Multiply by \(\frac{\sqrt[4]{y}}{\sqrt[4]{y}}\): \(\frac{\sqrt[4]{y}}{xy}\). 3. Multiply by \(\frac{\sqrt[m]{z}}{\sqrt[m]{z}}\): \(\frac{z\sqrt[m]{z}}{z}=\sqrt[m]{z}\).

1) \(\frac{u\sqrt[n]{v^{n-1}}}{v}\) 2) \(\frac{\sqrt[4]{y}}{xy}\) 3) \(\sqrt[m]{z}\)

- View \(1-a\) as a difference of squares. - Factor the numerator so one factor matches the radicand in the denominator. - Use \(u/\sqrt{u}=\sqrt{u}\) when \(u>0\).

1. Factor the numerator as a difference of squares: \(1-a=(1-\sqrt{a})(1+\sqrt{a})\). 2. Then \(T=\frac{(1-\sqrt{a})(1+\sqrt{a})}{\sqrt{1+\sqrt{a}}}\). 3. Since \(1+\sqrt{a}>0\), \(\frac{1+\sqrt{a}}{\sqrt{1+\sqrt{a}}}=\sqrt{1+\sqrt{a}}\). 4. Therefore, \(T=(1-\sqrt{a})\sqrt{1+\sqrt{a}}\).

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

- Multiply each denominator by its conjugate. - Apply the difference-of-squares identity in the denominator. - In b), simplify the two fractions before adding them.

1. In a), multiply by \(\frac{\sqrt{5}+2}{\sqrt{5}+2}\). The denominator is \(5-4=1\), and the numerator is \((\sqrt{5}+2)^2=9+4\sqrt{5}\). 2. In b), rationalize each fraction: \(\frac{1}{\sqrt{2}+1}=\sqrt{2}-1\) and \(\frac{1}{\sqrt{2}-1}=\sqrt{2}+1\). 3. Adding gives \((\sqrt{2}-1)+(\sqrt{2}+1)=2\sqrt{2}\).

a) \(9+4\sqrt{5}\) b) \(2\sqrt{2}\)

- Multiply a single radical denominator by itself. - Factor a difference of squares after rationalizing. - Use a conjugate for a denominator that is a difference of radicals.

1. Multiply by \(\frac{\sqrt{x^2-9}}{\sqrt{x^2-9}}\): \(\frac{(x-3)\sqrt{x^2-9}}{x^2-9}\). Since \(x^2-9=(x-3)(x+3)\) and \(x\ne3\), canceling gives \(\frac{\sqrt{x^2-9}}{x+3}\). 2. Multiply by the conjugate \(\sqrt{y+4}+\sqrt{y}\). The denominator becomes \((y+4)-y=4\), so the result is \(\sqrt{y+4}+\sqrt{y}\).

1) \(\frac{\sqrt{x^2-9}}{x+3}\) 2) \(\sqrt{y+4}+\sqrt{y}\)

- Multiply each denominator by its conjugate. - In part 1, square the binomial numerator carefully. - Simplify the difference of squares in each denominator.

1. Multiply by \(\frac{\sqrt{u+v}+\sqrt{u-v}}{\sqrt{u+v}+\sqrt{u-v}}\). The denominator is \((u+v)-(u-v)=2v\). The numerator is \(2u+2\sqrt{u^2-v^2}\). Dividing by \(2\) gives \(\frac{u+\sqrt{u^2-v^2}}{v}\). 2. Multiply by \(\frac{\sqrt{x^2+y^2}+x}{\sqrt{x^2+y^2}+x}\). The denominator becomes \(x^2+y^2-x^2=y^2\). Since \(y\ne0\), canceling \(y^2\) gives \(\sqrt{x^2+y^2}+x\).

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

- Rationalize using the conjugate of the denominator. - Use a binomial identity to square the simplified value of \(A\). - Substitute consistently and combine like terms. - Watch the minus sign when subtracting \(2A\).

1. Multiply by the conjugate: \(A=\frac{4(\sqrt{5}+1)}{5-1}=\sqrt{5}+1\). 2. Then \(A^2=(\sqrt{5}+1)^2=6+2\sqrt{5}\). 3. Substitute: \(A^2-2A-4=(6+2\sqrt{5})-2(\sqrt{5}+1)-4=0\).

a) \(A=\sqrt{5}+1\) b) \(0\)

- Multiply by the conjugate of the denominator. - Use the product property when multiplying radicals. - Simplify radical products before canceling common factors. - Check whether the final numerator and denominator share a common numerical factor.

1. Multiply the numerator and denominator by the conjugate \(4\sqrt{3}-\sqrt{15}\). 2. The denominator becomes \((4\sqrt{3})^2-(\sqrt{15})^2=48-15=33\). 3. The numerator is \((2\sqrt{15}-3\sqrt{3})(4\sqrt{3}-\sqrt{15})=11\sqrt{45}-66=33\sqrt{5}-66\). 4. Divide every term by \(33\) to obtain \(\sqrt{5}-2\).

\(\sqrt{5}-2\)

- For each denominator, identify its conjugate. - The product of conjugates removes the radical term from the denominator. - After rationalizing both expressions, look for a convenient way to use their sum and product rather than expanding both squares separately.

1. For \(x\), multiply by the conjugate \(\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}+\sqrt{3}}\). The denominator becomes \(7-3=4\), and the numerator is \((\sqrt{7}+\sqrt{3})^2=10+2\sqrt{21}\). Thus, \(x=\frac{5+\sqrt{21}}{2}\). 2. For \(y\), multiply by the conjugate \(\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}-\sqrt{3}}\). The denominator again becomes \(4\), and the numerator is \((\sqrt{7}-\sqrt{3})^2=10-2\sqrt{21}\). Thus, \(y=\frac{5-\sqrt{21}}{2}\). 3. From these rationalized forms, \(x+y=5\) and \(xy=\frac{25-21}{4}=1\). 4. Then \(x^2+y^2=(x+y)^2-2xy=25-2=23\). 5. Therefore, \(x^2-4xy+y^2=23-4=19\).

a) \(x=\frac{5+\sqrt{21}}{2}\), \(y=\frac{5-\sqrt{21}}{2}\) b) \(19\)

- Rationalize each reciprocal before applying the square. - Use the fact that the two binomial denominators are conjugates. - Simplify the rationalized fractions before squaring. - Expand the two squared binomials and combine like terms.

1. Because \((\sqrt{5}-2)(\sqrt{5}+2)=1\), \(\frac{1}{\sqrt{5}-2}=\sqrt{5}+2\) and \(\frac{1}{\sqrt{5}+2}=\sqrt{5}-2\). 2. Therefore, the expression becomes \((\sqrt{5}+2)^2+(\sqrt{5}-2)^2\). 3. Expanding gives \((9+4\sqrt{5})+(9-4\sqrt{5})=18\).

\(18\)

- How could you multiply the denominator by a related binomial to remove the radical there? - Check the original denominator before canceling any factor. - In part b), substitute into the new expression and simplify the number under the square root first.

1. The original denominator is zero when \(\sqrt{x^2+1}=1\), which gives \(x=0\). Thus, require \(x\ne0\). 2. Multiply by the conjugate: \(T(x)=\frac{x(\sqrt{x^2+1}+1)}{(\sqrt{x^2+1}-1)(\sqrt{x^2+1}+1)}\). 3. The denominator becomes \(x^2\), so \(T(x)=\frac{\sqrt{x^2+1}+1}{x}\) for \(x\ne0\). 4. At \(x=\frac{4}{3}\), \(\sqrt{x^2+1}=\frac{5}{3}\), hence \(T\left(\frac{4}{3}\right)=\frac{\frac{5}{3}+1}{\frac{4}{3}}=2\).

a) \(T(x)=\frac{\sqrt{x^2+1}+1}{x}\), with \(x\ne0\). b) \(T\left(\frac{4}{3}\right)=2\).

- What two-term factor will turn the denominator into a difference of squares? - Check when that factor is nonzero before using it to multiply by one. - Compare the original and new denominators at the input with the open circle.

1. For \(x\ge0\) and \(x\ne1\), multiply \(f(x)\) by \(\frac{\sqrt{x}-1}{\sqrt{x}-1}\). The new denominator is \(x-1\), giving \(g(x)=\frac{\sqrt{x}-1}{x-1}\), with domain \(x\ge0\), \(x\ne1\). 2. The expression for \(g\) is undefined at \(x=1\); the graph therefore has an open circle at \((1,\frac{1}{2})\). The original \(f(1)=\frac{1}{2}\) is defined. 3. The functions agree where both are defined, but their domains differ. They are not the same function.

a) \(g(x)=\frac{\sqrt{x}-1}{x-1}\), with \(x\ge0\) and \(x\ne1\). b) The open circle is at \((1,\frac{1}{2})\), where \(g\) is undefined but \(f(1)=\frac{1}{2}\). No; the functions have different domains.

- Rationalize the denominator using its conjugate. - Since both values are positive, compare their squares. - Reduce the final comparison to nearby perfect squares.

1. Rationalize the denominator: \(\frac{1}{\sqrt{7}-\sqrt{5}}=\frac{\sqrt{7}+\sqrt{5}}{2}\). 2. Both values are positive, so compare their squares. The square of the first value is \(\left(\frac{\sqrt{7}+\sqrt{5}}{2}\right)^2=3+\frac{\sqrt{35}}{2}\), while \((\sqrt{6})^2=6\). 3. Since \(\sqrt{35}<6\), \(3+\frac{\sqrt{35}}{2}<6\). Therefore, \(\frac{1}{\sqrt{7}-\sqrt{5}}<\sqrt{6}\).

\(\frac{1}{\sqrt{7}-\sqrt{5}}<\sqrt{6}\)

- Simplify numerical factors under a radical before taking the square root. - Remember that \(\sqrt{u^2}=|u|\). - Determine the sign of each expression inside an absolute value. - Rationalize the resulting binomial denominator using its conjugate.

1. In a), simplify the numerical fraction: \(\sqrt{4(4-\sqrt{15})^2}=2|4-\sqrt{15}|\). 2. Since \(4=\sqrt{16}>\sqrt{15}\), \(4-\sqrt{15}>0\). Thus, the result is \(2(4-\sqrt{15})=8-2\sqrt{15}\). 3. In b), \(\sqrt{\frac{3}{(2-\sqrt{7})^2}}=\frac{\sqrt{3}}{|2-\sqrt{7}|}\). Since \(2<\sqrt{7}\), the denominator is \(\sqrt{7}-2\). 4. Rationalize using the conjugate: \(\frac{\sqrt{3}}{\sqrt{7}-2}\cdot\frac{\sqrt{7}+2}{\sqrt{7}+2}=\frac{\sqrt{21}+2\sqrt{3}}{3}\).

a) \(8-2\sqrt{15}\) b) \(\frac{\sqrt{21}+2\sqrt{3}}{3}\)

- An odd power preserves the negative sign. - Cube the factor outside the radical and the radical factor separately. - Extract perfect-square powers from the resulting radical. - Multiply by a suitable radical to remove the radical from the denominator.

1. Cube the factors: \(\left(-\frac{x}{y}\right)^3\left(\sqrt{\frac{2y}{x^3}}\right)^3=-\frac{x^3}{y^3}\sqrt{\frac{8y^3}{x^9}}\). 2. Extract perfect-square factors and cancel: \(-\frac{2\sqrt{2y}}{xy^2\sqrt{x}}\). 3. Rationalize the denominator by multiplying by \(\frac{\sqrt{x}}{\sqrt{x}}\): \(-\frac{2\sqrt{2xy}}{x^2y^2}\).

\(-\frac{2\sqrt{2xy}}{x^2y^2}\)

- Before using a conjugate as a factor of one, check whether it could be zero. - Can you factor the numerator as a difference of two squares? - Which input pair makes both terms in the original denominator zero?

1. When \(b\ne c\) and both roots are real, multiply by \(\frac{\sqrt{b}-\sqrt{c}}{\sqrt{b}-\sqrt{c}}\). The denominator product becomes \(b-c\), so \(A=\sqrt{b}-\sqrt{c}\). 2. When \(b=c>0\), the original fraction is \(\frac{0}{2\sqrt{b}}=0\), which agrees with \(\sqrt{b}-\sqrt{c}=0\). The conjugate quotient cannot be used in this case. 3. Alternatively, factor \(b-c=(\sqrt{b}-\sqrt{c})(\sqrt{b}+\sqrt{c})\). The second factor in the denominator is nonzero on the original domain, so cancellation gives \(A=\sqrt{b}-\sqrt{c}\) for every allowed input. 4. Real roots require \(b\ge0\), \(c\ge0\); their sum is zero only if both are zero. Thus \((b,c)\ne(0,0)\).

a) For \(b\ne c\), multiplying by the conjugate gives \(A=\frac{(b-c)(\sqrt{b}-\sqrt{c})}{b-c}=\sqrt{b}-\sqrt{c}\). For \(b=c>0\), the original fraction gives \(A=0\) directly. b) Since \(b-c=(\sqrt{b}-\sqrt{c})(\sqrt{b}+\sqrt{c})\), canceling the nonzero original denominator gives \(A=\sqrt{b}-\sqrt{c}\) throughout the original domain. c) \(b\ge0\), \(c\ge0\), and \((b,c)\ne(0,0)\).

- A conjugate product is a difference of two squares. - Check when the transformed denominator can become zero, even if the original denominator is not zero. - At the requested inputs, choose whichever valid expression makes the calculation simpler.

1. Real roots require \(-1\le x\le1\). Multiply numerator and denominator by \(\sqrt{1+x}-\sqrt{1-x}\). 2. The new denominator is \((1+x)-(1-x)=2x\), while the numerator is \(2-2\sqrt{1-x^2}\). For \(x\ne0\), this gives \(A=\frac{1-\sqrt{1-x^2}}{x}\). 3. At \(x=0\), the original fraction is \(\frac{1-1}{1+1}=0\). The rationalized form from part a) cannot be used at zero. 4. At \(x=\frac{1}{2}\), the rationalized form gives \(A=\frac{1-\sqrt{\frac{3}{4}}}{\frac{1}{2}}=2-\sqrt{3}\).

a) \(A=\frac{1-\sqrt{1-x^2}}{x}\) for \(-1\le x\le1\), \(x\ne0\). b) \(A(0)=0\) and \(A\left(\frac{1}{2}\right)=2-\sqrt{3}\).