- The radical is already isolated. - Use an inverse operation that removes the square root. - Check the result in the original equation.

1. Square both sides: \(x+5=16\). 2. Solve: \(x=11\). 3. Substitution gives \(\sqrt{16}=4\), so the solution is valid.

\(x=11\)

- Match the inverse power to the radical index. - Compare the sign behavior of odd-index roots with square roots. - Verify the result in the original equation.

1. Cube both sides: \(x-1=-8\). 2. Solve: \(x=-7\). 3. A cube root can be negative because odd-index roots are defined for negative radicands. 4. Substitution gives \(\sqrt[3]{-8}=-2\).

\(x=-7\). A negative right side is allowed because cube roots can have negative real values.

- Consider the possible output values of a principal square root. - Decide whether the right side is in that output range.

1. A principal square root is always nonnegative wherever it is real. 2. Therefore, \(\sqrt{2x+3}\) cannot equal \(-1\) for any real \(x\).

No. The equation has no real solution because a principal square root cannot be negative.

- Use the original equation, not the squared equation, for the check. - Compare the two sides after substitution.

1. Substitute \(x=0\) into the original equation. 2. The left side is \(\sqrt{1}=1\), while the right side is \(-1\). 3. Since the two sides are not equal, \(x=0\) is extraneous.

\(x=0\) is not a solution; it is extraneous.

- Which term is outside the square root? - What inverse operation will leave the square root alone on one side?

1. Subtract \(2\) from both sides to isolate the radical: \(\sqrt{x+5}=4\).

\(\sqrt{x+5}=4\)

- Think about the sign of a principal square root. - Does an odd root follow the same sign rule?

1. A principal square root cannot be negative, so a) has no real solution. 2. A cube root can be negative. In b), the equation holds when \(x+1=-1\), so \(x=-2\).

b) only; it has solution \(x=-2\). A principal square root cannot equal \(-1\).

- What operation undoes a square root when the radical is isolated? - Apply the same operation to the entire left and right sides.

1. Squaring both sides gives \((\sqrt{x+4})^2=5^2\). 2. The resulting equation is \(x+4=25\).

\(x+4=25\)

- What does the principal square-root symbol mean? - Can two unequal numbers have the same square?

1. The principal square root is \(\sqrt{9}=3\), not \(-3\). Thus \(\sqrt{9}=-3\) is false. 2. Both sides of this false statement have square \(9\). Squaring can therefore turn unequal numbers into equal squares. Answers found after squaring must be checked in the original equation.

\(\sqrt{9}=-3\) is false. Numbers with opposite signs have equal squares, so squaring can create extra candidates that must be checked.

- Find the least possible value of \(x^2\). - Use that value to find the least possible radicand. - Use the principal square root to find the least possible value of the left side. - Compare the least possible value of the left side with \(4\).

1. For every real \(x\), \(x^2\ge0\), so \(x^2+25\ge25\). 2. Therefore, \(\sqrt{x^2+25}\ge\sqrt{25}=5\). 3. The left side can never equal \(4\), so the equation has no real solution.

There is no real solution because \(\sqrt{x^2+25}\ge5\) for every real \(x\), but the equation requires it to equal \(4\).

- Recall the range of the principal square-root function. - Determine the domain conditions for every radical. - In c), isolate the radical and inspect the required value. - Check whether the separate domain conditions overlap.

1. In a), a principal square root is always nonnegative, so it cannot equal \(-5\). 2. In b), the first radical requires \(x\ge8\), while the second requires \(x\le2\). No real number satisfies both conditions, so the equation has no real domain. 3. In c), isolating the radical would give \(\sqrt{x-1}=-7\), which is impossible because a principal square root is nonnegative.

a) No real solution; a square root cannot equal \(-5\). b) No real solution; the conditions \(x\ge8\) and \(x\le2\) cannot both hold. c) No real solution; the equation would require \(\sqrt{x-1}=-7\).

- What operation would undo the outer operation in each equation? - After removing the outer operation, what type of equation remains? - Check each proposed value in the original equation, especially where a square root appears.

1. For a), cube both sides: \(3x+10=4^3=64\), so \(x=18\). 2. For b), square both sides: \(2x-5=x+1\), so \(x=6\). 3. Check b) in the original equation: both radicands are \(7\), so \(x=6\) is valid.

a) \(\{18\}\) b) \(\{6\}\)

- Square both sides, then solve the resulting quadratic equation. - A principal square root cannot be negative. - Check each candidate in the original equation.

1. Square both sides: \(x+7=(x+1)^2\). 2. Expand and rearrange: \(x^2+x-6=0\). 3. Factor: \((x+3)(x-2)=0\), so the candidates are \(x=-3\) and \(x=2\). 4. Check the original equation. For \(x=2\), both sides equal \(3\). For \(x=-3\), the left side is \(2\) and the right side is \(-2\), so \(-3\) is extraneous.

\(\{2\}\)

- Find the common domain of both radicals. - Square both sides to obtain a quadratic equation. - Check the candidates against the domain.

1. The domain requires \(x\ge1\). 2. Square both sides: \((x+2)(x-1)=10\). 3. Expand and rearrange: \(x^2+x-12=0\). 4. Factor: \((x+4)(x-3)=0\). The candidates are \(x=-4\) and \(x=3\). 5. Only \(x=3\) is in the domain, and \(\sqrt{5}\cdot\sqrt{2}=\sqrt{10}\).

\(\{3\}\)

- Treat the repeated radical expression as one quantity and collect like terms. - Isolate the square root before squaring. - Subtract fractions using a common denominator when needed. - Check the resulting value in the original equation.

1. Collect the radical terms on one side and the constants on the other: \(\frac{7}{4}\sqrt{2x}-\frac{3}{4}\sqrt{2x}=13-5\). 2. This simplifies to \(\sqrt{2x}=8\). 3. Square both sides: \(2x=64\), so \(x=32\). 4. Substitution confirms that \(x=32\) satisfies the original equation.

\(\{32\}\)

- Square both sides when the isolated square root equals a nonnegative value. - Remember that solving \(x^2=a\) may produce two real values. - A principal square root cannot equal a negative number.

1. For part a, square both sides: \(2x+7=25\). Thus \(x=9\), and substitution confirms the solution. 2. For part b, square both sides: \(x^2-9=16\), so \(x^2=25\) and \(x=\pm5\). Both values satisfy the original equation. 3. For part c, a principal square root is never negative, so the equation has no real solution.

a) \(\{9\}\) b) \(\{-5,5\}\) c) \(\emptyset\)

- List the zeros of the individual factors. - Find the intersection of the three radical-domain conditions. - Keep only candidate zeros that belong to the common domain.

1. The factors are individually zero at \(x=0\), \(x=-3\), and \(x=2\). 2. All three radicals must be defined at the same time. Their conditions are \(x\ge0\), \(x\ge-3\), and \(x\le2\), so the common domain is \([0, 2]\). 3. Of the three candidate values, only \(0\) and \(2\) lie in the domain. Both make the product zero. 4. Therefore, the equation has two real solutions.

\(2\) real solutions: \(x=0\) and \(x=2\)

- Find the common domain of all three radicals. - List the zeros of the individual factors. - Check which candidate zeros belong to the common domain.

1. The individual factors are zero at \(x=1\), \(x=3\), and \(x=5\). 2. For all three radicals to be defined, \(x\ge1\), \(x\ge3\), and \(x\ge5\) must hold. The common domain is \([5, \infty)\). 3. Only \(x=5\) is both a candidate zero and an element of the domain. 4. Therefore, the equation has one real solution.

\(1\) real solution: \(x=5\)

- Find the common domain from the two radical conditions. - List the zeros supplied by all three factors. - Count the candidate zeros that lie in the domain.

1. The factors give the candidate zeros \(x=4\), \(x=6\), and \(x=2\). 2. The radical conditions are \(x\le6\) and \(x\ge2\), so the domain is \([2, 6]\). 3. All three candidate zeros lie in the domain and make the product zero. 4. Therefore, the equation has three real solutions.

\(3\) real solutions: \(x=2,4,6\)

- The radical is already isolated. - Remove the square root with an inverse operation before solving the squared expression. - Remember that a positive square can come from two opposite values.

1. Square both sides: \((x-2)^2+4=36\). 2. Subtract \(4\): \((x-2)^2=32\). 3. Therefore, \(x-2=\pm\sqrt{32}=\pm4\sqrt{2}\), so \(x=2\pm4\sqrt{2}\). 4. Both values make the original distance equal to \(6\,\text{mi}\).

\(x=2-4\sqrt{2}\) or \(x=2+4\sqrt{2}\)

- Which power undoes a cube root? - After removing the cube root, how can you isolate \(x\)?

1. Cube both sides: \(2x-1=27\), so \(x=14\). 2. At \(x=14\), \(\sqrt[3]{27}=3\), so the result checks.

\(x=14\). Check: \(\sqrt[3]{2(14)-1}=\sqrt[3]{27}=3\).

- An original solution must be an intersection of the two plotted graphs. - Check whether the two sides actually have equal values at each candidate, not just equal squares.

1. At \(x=3\), the radical curve and the line meet at \((3,2)\), so the original equation holds. 2. At \(x=0\), the radical has value \(1\), while the line has value \(-1\); the sides have equal squares but different signs. 3. Thus \(x=0\) is extraneous and \(x=3\) is valid.

\(x=3\) is the solution. Reject \(x=0\): the square-root side is \(1\), while the line is \(-1\), so the original equation is false there.

- Substitute \(x=2\) into the original radicals. - Write the nonnegative-radicand condition for each side. - Check the domain before accepting a value obtained after squaring.

1. The left radical requires \(x-3\ge0\), so \(x\ge3\). 2. The right radical requires \(1-x\ge0\), so \(x\le1\). 3. No real number satisfies both conditions, so the original equation has an empty domain. 4. Therefore, \(x=2\) is not a solution. Substitution would produce \(\sqrt{-1}\), which is not real.

\(x=2\) is not a solution. The original equation has no real domain because it requires both \(x\ge3\) and \(x\le1\).

- Work from the outermost operation inward. - Try two different signs for a number with a positive square. What happens? - Can you verify possible solutions directly in the original equations?

1. For a), square both sides: \(11+\sqrt[3]{x-5}=9\). Thus \(\sqrt[3]{x-5}=-2\); cubing gives \(x=-3\). The original then becomes \(\sqrt9=3\), so the solution works. 2. For b), write \((x-4)^{\frac{2}{3}}=\left(\sqrt[3]{x-4}\right)^2\). Hence \(\sqrt[3]{x-4}=\pm2\), giving \(x-4=8\) or \(x-4=-8\), so \(x=12\) or \(x=-4\). 3. Both signs work because \(2^2=(-2)^2=4\); substitution confirms both original solutions.

a) \(x=-3\) b) \(\{-4,12\}\). Both positive and negative cube-root values square to \(4\).

- Find where both radicands are nonnegative and both denominators are nonzero. - Square both sides to remove the radicals. - Solve the resulting rational equation, then check the answer in the original equation.

1. Determine the domain. The left radicand is nonnegative for \(x\in(-\infty, -1]\cup(3, \infty)\), and the right radicand is nonnegative for \(x\in(-\infty, -5]\cup(1, \infty)\). Their intersection is \(D=(-\infty, -5]\cup(3, \infty)\). 2. Square both sides: \(\frac{x+1}{x-3}=\frac{x+5}{x-1}\). 3. Cross-multiply: \((x+1)(x-1)=(x+5)(x-3)\). 4. Expand and solve: \(x^2-1=x^2+2x-15\), so \(14=2x\) and \(x=7\). 5. Since \(7\in D\), check it in the original equation: both sides equal \(\sqrt{2}\).

\(\{7\}\)

- Account for every square root and denominator when finding the domain. - Square the entire equation. - Clear the fractions and solve the resulting quadratic equation. - Check both candidates in the original equation.

1. Determine the domain. The denominator \(\sqrt{x+1}\) requires \(x>-1\), and \(\frac{x}{x-1}\ge0\) requires \(x\in(-\infty, 0]\cup(1, \infty)\). Thus \(D=(-1, 0]\cup(1, \infty)\). 2. Square both sides: \(\frac{x}{x-1}=\frac{6}{x+1}\). 3. Cross-multiply: \(x(x+1)=6(x-1)\). 4. Expand and rearrange: \(x^2-5x+6=0\). Factor: \((x-2)(x-3)=0\), so \(x=2\) or \(x=3\). 5. Both values are in the domain and satisfy the original equation.

\(\{2,3\}\)

- Square both sides, then solve the resulting quadratic equation. - Remember that a principal square root is never negative. - Substitute every candidate into the original equation.

1. Square both sides: \(2x^2-7=(x+1)^2\). 2. Expand and rearrange: \(2x^2-7=x^2+2x+1\), so \(x^2-2x-8=0\). 3. Factor: \((x-4)(x+2)=0\). The candidates are \(x=4\) and \(x=-2\). 4. Check \(x=4\): \(\sqrt{2(4)^2-7}=5\) and \(4+1=5\), so it works. 5. Check \(x=-2\): \(\sqrt{2(-2)^2-7}=1\), but \(-2+1=-1\), so it is extraneous. 6. Squaring can make unequal expressions equal because it removes sign information, so every candidate must be checked.

\(\{4\}\). Checking is necessary because squaring can introduce extraneous solutions.

- A square root in a denominator must be strictly positive. - Clear the radical denominator before squaring. - After solving the quadratic equation, check the sign condition and the original equation.

1. The radicands require \(5x-1\ge0\) and \(x+2\ge0\). Because \(\sqrt{x+2}\) is in the denominator, \(x+2>0\). Together, these conditions give \(D=[0.2, \infty)\). 2. Multiply by \(\sqrt{x+2}\): \(\sqrt{(5x-1)(x+2)}=4x-2\). 3. Square both sides: \((5x-1)(x+2)=(4x-2)^2\). 4. Expand and rearrange: \(5x^2+9x-2=16x^2-16x+4\), so \(11x^2-25x+6=0\). 5. Use the quadratic formula: \(x=\frac{25\pm\sqrt{361}}{22}=\frac{25\pm19}{22}\). The candidates are \(x=2\) and \(x=\frac{3}{11}\). 6. The left side after step 2 is nonnegative, so \(4x-2\) must also be nonnegative. The candidate \(x=\frac{3}{11}\) fails this condition and is extraneous. The value \(x=2\) satisfies the original equation.

a) \(D=[0.2, \infty)\) b) \(\{2\}\)

- Find the domain first. - Isolate one radical before each squaring step. - Use the binomial formula carefully when squaring a difference. - Substitute every candidate into the original equation.

1. The domain requires \(x\ge-1\). 2. Isolate one radical: \(\sqrt{2x+3}=5-\sqrt{x+1}\). 3. Square and simplify: \(2x+3=25-10\sqrt{x+1}+x+1\), so \(x-23=-10\sqrt{x+1}\). 4. Square again: \((x-23)^2=100(x+1)\). This gives \(x^2-146x+429=0\). 5. Solve the quadratic: \(x=73\pm70\), so the candidates are \(x=3\) and \(x=143\). 6. Check the original equation. For \(x=3\), \(\sqrt{4}+\sqrt{9}=5\). For \(x=143\), \(\sqrt{144}+\sqrt{289}=29\ne5\). Therefore, \(143\) is extraneous. 7. Squaring is not reversible without sign restrictions, so it can introduce extraneous solutions.

\(\{3\}\). The candidate \(143\) is extraneous, which shows why every candidate must be checked.

- Isolate one radical before squaring. - Simplify the equation before the second squaring step. - Check every candidate in the original equation.

1. The domain requires \(x\ge2\). 2. Isolate one radical: \(\sqrt{4x+1}=3+\sqrt{x-2}\). 3. Square and simplify: \(4x+1=x+7+6\sqrt{x-2}\), so \(x-2=2\sqrt{x-2}\). 4. Square again: \((x-2)^2=4(x-2)\). This gives \(x^2-8x+12=0\). 5. Factor: \((x-2)(x-6)=0\), so the candidates are \(x=2\) and \(x=6\). 6. Check: for \(x=2\), \(\sqrt{9}-0=3\); for \(x=6\), \(\sqrt{25}-\sqrt{4}=3\). Both are valid.

\(\{2,6\}\)

- A square root in a denominator must be strictly positive. - Multiply by the radical denominator to simplify the equation. - Check the candidates against the domain and the original equation.

1. Since \(\sqrt{x+1}\) is in a denominator, the domain requires \(x+1>0\). The other radicand requires \(2x+8\ge0\). Thus \(D=(-1, \infty)\). 2. Multiply by \(\sqrt{x+1}\): \(x+1+3=\sqrt{(2x+8)(x+1)}\), so \(x+4=\sqrt{2x^2+10x+8}\). 3. Square both sides: \((x+4)^2=2x^2+10x+8\). 4. Expand and rearrange: \(x^2+2x-8=0\). Factor: \((x+4)(x-2)=0\), so the candidates are \(x=-4\) and \(x=2\). 5. The value \(-4\) is outside the domain. The value \(2\) satisfies the original equation.

\(\{2\}\)

- Determine the domain before multiplying by the radical denominator. - After isolating a square root, the other side must be nonnegative. - Check the candidates after solving the quadratic equation.

1. The denominator requires \(5x-1>0\), and \(\sqrt{x}\) requires \(x\ge0\). Thus \(D=(0.2, \infty)\). 2. Multiply by \(\sqrt{5x-1}\): \(5x-3=\sqrt{x(5x-1)}=\sqrt{5x^2-x}\). 3. Because the right side is nonnegative, require \(5x-3\ge0\), or \(x\ge0.6\). Square both sides: \((5x-3)^2=5x^2-x\). 4. Expand and rearrange: \(20x^2-29x+9=0\). 5. Use the quadratic formula: \(x=\frac{29\pm\sqrt{121}}{40}=\frac{29\pm11}{40}\). The candidates are \(x=1\) and \(x=0.45\). 6. The value \(0.45\) fails the condition \(x\ge0.6\) and is extraneous. The value \(1\) satisfies the original equation.

\(\{1\}\)

- Use a common denominator on the left side. - The denominator is a product of conjugates. - Simplify completely before solving the resulting polynomial equation.

1. Since \(x^2+8>0\) and \(\sqrt{x^2+8}>|x|\), neither denominator is zero. The domain is all real numbers. 2. Combine the fractions. The common denominator is \((\sqrt{x^2+8}-x)(\sqrt{x^2+8}+x)=x^2+8-x^2=8\). 3. The combined numerator is \(4(\sqrt{x^2+8}+x)-4(\sqrt{x^2+8}-x)=8x\). 4. The equation simplifies to \(\frac{8x}{8}=x^2\), or \(x=x^2\). 5. Factor: \(x(x-1)=0\). Thus \(x=0\) or \(x=1\), and both satisfy the original equation.

\(\{0,1\}\)

- Isolate one radical before squaring. - If a radical remains, isolate it and square again. - Check all candidates in the original equation.

1. The domain requires \(x\ge3\). 2. Isolate one radical: \(\sqrt{3x+4}=3+\sqrt{x-3}\). 3. Square and simplify: \(3x+4=x+6+6\sqrt{x-3}\), so \(x-1=3\sqrt{x-3}\). 4. Square again: \((x-1)^2=9(x-3)\). This gives \(x^2-11x+28=0\). 5. Factor: \((x-4)(x-7)=0\), so the candidates are \(x=4\) and \(x=7\). 6. Both satisfy the original equation: \(4-1=3\) when \(x=4\), and \(5-2=3\) when \(x=7\).

\(\{4,7\}\)

- Find the common domain of all three radical expressions. - Square carefully and include the cross-product term. - Check candidates against the domain and the original equation.

1. The domain requires \(x\ge3\). 2. Square both sides: \(x+(x-3)+2\sqrt{x(x-3)}=2x+1\). 3. Simplify: \(2\sqrt{x^2-3x}=4\), so \(\sqrt{x^2-3x}=2\). 4. Square again: \(x^2-3x=4\), or \(x^2-3x-4=0\). 5. Factor: \((x-4)(x+1)=0\). The candidates are \(x=4\) and \(x=-1\). 6. Only \(x=4\) is in the domain, and substitution gives \(2+1=3\).

\(\{4\}\)

- Find the domain before squaring. - Rearrange so one side is a sum of nonnegative radical terms. - When a radical is isolated, the other side must also be nonnegative. - Check every candidate.

1. The domain requires \(x\ge2.5\). 2. Rewrite the equation as \(\sqrt{x+6}=\sqrt{2x-5}+\sqrt{x+1}\). 3. Square and isolate the remaining radical: \(x+6=3x-4+2\sqrt{(2x-5)(x+1)}\), so \(5-x=\sqrt{2x^2-3x-5}\). This also requires \(x\le5\). 4. Square again: \((5-x)^2=2x^2-3x-5\). Rearranging gives \(x^2+7x-30=0\). 5. Factor: \((x-3)(x+10)=0\). The candidates are \(x=3\) and \(x=-10\). 6. Only \(x=3\) satisfies the domain and sign conditions. Substitution gives \(3-2=1\).

\(\{3\}\)

- Multiply by the radical denominator first. - Isolate the remaining radical before squaring. - Record the sign condition created when the radical is isolated. - Check the result in the original equation.

1. Because \(\sqrt{x-3}\) is in the denominator, the domain is \(x>3\). 2. Multiply by \(\sqrt{x-3}\): \(x-3+\sqrt{(x+5)(x-3)}=8\). 3. Isolate the radical: \(\sqrt{x^2+2x-15}=11-x\). This requires \(x\le11\). 4. Square: \(x^2+2x-15=(11-x)^2\). Simplifying gives \(24x=136\), so \(x=\frac{17}{3}\). 5. This value is in the domain and satisfies the original equation.

\(\left\{\frac{17}{3}\right\}\)

- A radical denominator must be strictly positive. - Multiply by the denominator before isolating and squaring the remaining radical. - Check candidates against the domain and the original equation.

1. The denominator requires \(x+1>0\), and the other radicand requires \(8x+1\ge0\). Thus \(D=[-0.125, \infty)\). 2. Multiply by \(\sqrt{x+1}\): \(x+7=\sqrt{(8x+1)(x+1)}=\sqrt{8x^2+9x+1}\). 3. Square both sides: \((x+7)^2=8x^2+9x+1\), which simplifies to \(7x^2-5x-48=0\). 4. Use the quadratic formula: \(x=\frac{5\pm37}{14}\). The candidates are \(x=3\) and \(x=-\frac{16}{7}\). 5. The value \(-\frac{16}{7}\) is outside the domain. The value \(3\) satisfies the original equation.

a) \(D=[-0.125, \infty)\) b) \(x=3\) and \(x=-\frac{16}{7}\) c) \(\{3\}\)

- A square root in a denominator must be strictly positive. - Clear the denominator, then square both sides. - Check the quadratic solutions against the domain.

1. Because \(\sqrt{x-1}\) is in the denominator, the domain is \(x>1\). 2. Multiply by \(\sqrt{x-1}\): \(\sqrt{(3x+1)(x-1)}=x+3\). 3. Square both sides: \((3x+1)(x-1)=(x+3)^2\). 4. Expand and simplify: \(3x^2-2x-1=x^2+6x+9\), so \(x^2-4x-5=0\). 5. Factor: \((x-5)(x+1)=0\). The candidates are \(x=5\) and \(x=-1\). 6. Only \(x=5\) is in the domain, and substitution gives \(4=\frac{8}{2}\).

\(\{5\}\)

- Find the domain first. - When squaring a sum, include the cross-product term. - Isolate the remaining radical before squaring again. - Check both candidates.

1. The domain requires \(x\ge-1\). 2. Square both sides: \(x+1+x+6+2\sqrt{(x+1)(x+6)}=5x+10\). 3. Isolate the remaining radical: \(2\sqrt{x^2+7x+6}=3x+3\). 4. Square again: \(4(x^2+7x+6)=(3x+3)^2\). Simplifying gives \(x^2-2x-3=0\). 5. Factor: \((x-3)(x+1)=0\), so the candidates are \(x=3\) and \(x=-1\). 6. Both satisfy the original equation: \(2+3=5\) for \(x=3\), and \(0+\sqrt{5}=\sqrt{5}\) for \(x=-1\).

\(\{-1,3\}\)

- Find the common domain first. - Square the sum carefully, including the cross-product term. - After isolating the remaining radical, compare the signs of both sides.

1. The common domain is \(x\ge-0.5\). 2. Square both sides: \(x+3+x+8+2\sqrt{(x+3)(x+8)}=2x+1\). 3. Simplify: \(2\sqrt{x^2+11x+24}=-10\), or \(\sqrt{x^2+11x+24}=-5\). 4. A principal square root cannot be negative, so the equation has no real solution.

\(\emptyset\)

- Extract perfect-square factors from \(12\) and \(75\). - Combine the resulting like radical terms before squaring. - Isolate the remaining radical before squaring both sides. - Verify the solution in the original equation.

1. Simplify the radicals: \(\sqrt{12x}=2\sqrt{3x}\) and \(\sqrt{75x}=5\sqrt{3x}\). 2. Combine like radical terms: \(7\sqrt{3x}=14\), so \(\sqrt{3x}=2\). 3. Square both sides: \(3x=4\), giving \(x=\frac{4}{3}\). 4. The value is nonnegative and satisfies the original equation.

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

- A radicand of a real square root must be nonnegative. - For the second equation, compare the domain condition from each side. - For the first equation, square both sides and then check the radicands.

1. Square both sides of the first equation: \(4x-5=x+7\). Thus \(3x=12\) and \(x=4\). Both radicands equal \(11\), so the solution is valid. 2. For the second equation, \(\sqrt{x-10}\) requires \(x\ge10\), while \(\sqrt{4-x}\) requires \(x\le4\). No real number satisfies both conditions, so the common domain is empty.

1. \(x=4\) 2. No real solution; the domain conditions \(x\ge10\) and \(x\le4\) cannot both hold.

- Clear the radical denominator first. - Isolate the remaining radical before squaring. - Record the sign condition on the other side and check the result.

1. The domain requires \(x\ge-1\). 2. Multiply by \(\sqrt{x+6}\): \(\sqrt{(x+1)(x+6)}+x+6=15\). 3. Isolate the radical: \(\sqrt{x^2+7x+6}=9-x\). This requires \(x\le9\). 4. Square both sides: \(x^2+7x+6=(9-x)^2\). Simplifying gives \(25x=75\), so \(x=3\). 5. The value \(3\) is in the domain and satisfies the original equation.

\(\{3\}\)

- Find the domain before clearing the radical denominator. - Isolate one radical before squaring. - Check the result in the original equation.

1. The denominator requires \(x+3>0\), and the other radicand requires \(4x+1\ge0\). Thus \(D=[-0.25, \infty)\). 2. Multiply by \(\sqrt{x+3}\): \(2(x+3)-\sqrt{(4x+1)(x+3)}=3\). 3. Isolate the radical: \(\sqrt{4x^2+13x+3}=2x+3\). 4. Square both sides: \(4x^2+13x+3=(2x+3)^2\). Simplifying gives \(x=6\). 5. Substitution yields \(2\cdot3-5=\frac{3}{3}\), so \(x=6\) is valid.

\(\{6\}\)

- A real square root requires a nonnegative radicand. - Read \(f(2)\) from the graph before applying the square root. - At an intersection, solve \(y=\sqrt{y}\) for possible function values \(y\). - Then read where \(f(x)\) has those values.

1. The function \(g\) is defined where \(f(x)\ge0\). From the graph, \(f(4)<0\), so \(4\) is not in the domain of \(g\). 2. The graph shows \(f(2)=1\). Therefore, \(g(2)=\sqrt{1}=1\). 3. At an intersection, \(f(x)=\sqrt{f(x)}\), with \(f(x)\ge0\). Let \(y=f(x)\). Then \(y=\sqrt{y}\), so \(y^2=y\), which gives \(y=0\) or \(y=1\). 4. From the graph, \(f(x)=0\) at \(x=-2, -1, 3\), and \(f(x)=1\) at \(x=-3, 1, 2\). 5. Thus, the x-coordinates of the intersections are \(-3, -2, -1, 1, 2, 3\).

a) \(f(4)<0\), so \(4\) is not in the domain of \(g\). Also, \(g(2)=1\). b) \(x\in\{-3, -2, -1, 1, 2, 3\}\)

- Determine where the graph of \(f\) is on or above the x-axis. - Read \(f(1)\) before applying the square root. - At a shared point, solve \(y=\sqrt{y}\) for possible output values.

1. The square root is defined where \(f(x)\ge0\). The graph is on or above the x-axis for \(0\le x\le2\), so the domain of \(h\) is \([0, 2]\). 2. The graph shows \(f(1)=1\), so \(h(1)=\sqrt{1}=1\). 3. At a shared point, \(f(x)=\sqrt{f(x)}\), with \(f(x)\ge0\). Let \(y=f(x)\). Then \(y=\sqrt{y}\), so \(y^2=y\), giving \(y=0\) or \(y=1\). 4. From the graph, \(f(x)=0\) at \(x=0, 2\), and \(f(x)=1\) at \(x=1\). Therefore, the shared points have x-coordinates \(0, 1, 2\).

a) \([0, 2]\) b) \(h(1)=1\) c) \(x=0, 1, 2\)

- Square only after checking that the radical is isolated. - Simplify the resulting polynomial equation completely. - Use the stated physical condition to reject any impossible length. - Verify the remaining value in the original length equation.

1. Square both sides: \(x^2+(x+2)^2=(x+5)^2\). 2. Expand and simplify: \(2x^2+4x+4=x^2+10x+25\), so \(x^2-6x-21=0\). 3. Apply the quadratic formula: \(x=\frac{6\pm\sqrt{36+84}}{2}=3\pm\sqrt{30}\). 4. The physical condition \(x>0\) leaves \(x=3+\sqrt{30}\). 5. The side lengths are \((3+\sqrt{30})\,\text{ft}\), \((5+\sqrt{30})\,\text{ft}\), and \((8+\sqrt{30})\,\text{ft}\), which satisfy the original equation.

\(x=3+\sqrt{30}\). The side lengths are \((3+\sqrt{30})\,\text{ft}\), \((5+\sqrt{30})\,\text{ft}\), and \((8+\sqrt{30})\,\text{ft}\).

- First impose the condition that the proposed x-value works after squaring. - Then return to the unsquared equation. - Compare the parameter choices by checking the sign of the original right side.

1. For \(x=0\) to satisfy the squared equation, \(4=a^2\), so \(a=2\) or \(a=-2\). 2. In the original equation at \(x=0\), the left side is \(2\) and the right side is \(-a\). 3. If \(a=-2\), the original equation reads \(2=2\), so \(x=0\) is valid. 4. If \(a=2\), the original equation reads \(2=-2\), so \(x=0\) is extraneous.

\(a=2\)

- Before squaring, what sign must the right-hand side have? - Squaring may give candidates that were not solutions of the original equation. - Test every candidate in the unsquared equation.

1. Squaring gives \(x+4=(x-2)^2=x^2-4x+4\), so \(x(x-5)=0\). 2. The candidates are \(x=0\) and \(x=5\). At \(x=0\), the sides are \(2\) and \(-2\), so reject it. 3. At \(x=5\), both sides are \(3\), so it is valid.

\(x=5\). At \(x=0\), the sides are \(2\) and \(-2\), so reject it. At \(x=5\), both sides are \(3\).

- How do the cable, post, and ground form a right triangle? - Write the cable length in two ways: from the triangle and from the stated difference. - Check the ground distance in the original length equation, before squaring.

1. Let \(x\ge0\) be the ground distance in feet. The right triangle gives cable length \(\sqrt{x^2+64}\), and the stated length difference gives \(x+2\). 2. Set \(\sqrt{x^2+64}=x+2\). Squaring gives \(x^2+64=(x+2)^2=x^2+4x+4\). 3. Then \(60=4x\), so \(x=15\,\text{ft}\). 4. Check: \(\sqrt{15^2+64}=17=15+2\).

\(\sqrt{x^2+64}=x+2\), with \(x\ge0\). The ground distance is \(15\,\text{ft}\). Check: \(\sqrt{15^2+64}=17=15+2\).

- Find the interval where both radicands are nonnegative, then exclude values that make a denominator zero. - Combine the fractions using the product of conjugates. - Simplify before solving the remaining radical equation.

1. The radicands require \(-1\le x\le1\). The first denominator is zero when \(x=0\), so \(D=[-1, 1]\setminus\{0\}\). 2. Combine the fractions. The common denominator is \((\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})=2x\). 3. The combined numerator is \(x(\sqrt{1+x}+\sqrt{1-x})-x(\sqrt{1+x}-\sqrt{1-x})=2x\sqrt{1-x}\). 4. Since \(x\ne0\), the left side simplifies to \(\sqrt{1-x}\). Thus \(\sqrt{1-x}=\frac{1}{2}\). 5. Square both sides: \(1-x=\frac{1}{4}\), so \(x=\frac{3}{4}\). This value is in the domain and satisfies the original equation.

\(\left\{\frac{3}{4}\right\}\)

- Use the product equation to simplify \(\sqrt{xy}\). - Substitute new variables for \(\sqrt{x}\) and \(\sqrt{y}\). - Solve the resulting sum-and-product system, then square the values.

1. The denominators require \(x > 0\) and \(y > 0\). Therefore, \(\sqrt{xy} = 8\). 2. Combine the fractions: \(\frac{\sqrt{x} + \sqrt{y}}{\sqrt{xy}} = \frac{3}{4}\). Substitute \(\sqrt{xy} = 8\): \(\sqrt{x} + \sqrt{y} = 6\). 3. Let \(u = \sqrt{x}\) and \(v = \sqrt{y}\). Then \(u + v = 6\) and \(uv = 8\). 4. Substitute \(v = 6 - u\): \(u(6 - u) = 8\), so \(u^2 - 6u + 8 = 0\). 5. Factor: \((u - 2)(u - 4) = 0\). Thus, \((u, v) = (2, 4)\) or \((4, 2)\). 6. Since \(x = u^2\) and \(y = v^2\), the solutions are \((4, 16)\) and \((16, 4)\).

\(\{(4, 16), (16, 4)\}\)