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Algebra II : Solving Radical Equations

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #4277 : Algebra Ii

Solve: $\sqrt{8-x}+\sqrt{9-x} =2$

Possible Answers:

$\frac{53}{5}$

$\frac{13}{7}$

$\frac{119}{16}$

$\frac{99}{8}$

$\frac{131}{24}$

Correct answer:

$\frac{119}{16}$

Explanation:

Square both sides.

$(\sqrt{8-x}+\sqrt{9-x} )^2=2^2$

We will need to simplify the left side using the FOIL method.

$(\sqrt{8-x}+\sqrt{9-x} )(\sqrt{8-x}+\sqrt{9-x} )$

A radical multiplied by itself becomes what's inside the radical.

The expression becomes:

$(8-x)+(\sqrt{8-x})(\sqrt{9-x})+(\sqrt{9-x})(\sqrt{8-x})+(9-x)$

Combine like-terms and rewrite the equation.

$-2x+2\sqrt{8-x}\sqrt{9-x}+17 = 4$

Isolate the radical by adding and subtract 17 on both sides. Then divide by two on both sides.

$2\sqrt{8-x}\sqrt{9-x} =2x-13$

$\frac{2\sqrt{8-x}\sqrt{9-x} }{2}=\frac{2x-13}{2}$

$\sqrt{8-x}\sqrt{9-x} = x-\frac{13}{2}$

The radicals can be combined as one whole using the FOIL method.

$\sqrt{72-17x+x^2} = x-\frac{13}{2}$

Square both sides.

$(\sqrt{72-17x+x^2}) ^2= (x-\frac{13}{2})^2$

$72-17x+x^2 = x^2-13x+\frac{169}{4}$

The x^2 terms will cancel by subtracting the term on both side, and we can solve for .

$72-17x = -13x+\frac{169}{4}$

Add 17x on both sides.

$72 = 4x+\frac{169}{4}$

Multiply by four on both sides to eliminate the fraction.

$4[72 = 4x+\frac{169}{4}]$

288 = 16x+169

Solve for x.

288-169 = 16x+169-169

119 = 16x

The answer is: $\frac{119}{16}$

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Example Question #4278 : Algebra Ii

Solve: $4\sqrt{3x-3} = 12$

Possible Answers:

Correct answer:

Explanation:

Divide by four on both sides.

$\frac{4\sqrt{3x-3} }{4}= \frac{12}{4}$

The equation becomes:

$\sqrt{3x-3} =3$

Square both sides.

$(\sqrt{3x-3}) ^2=3^2$

3x-3= 9

Add three on both sides.

3x-3+3= 9+3

3x =12

Divide by three on both sides.

$\frac{3x}{3} =\frac{12}{3}$

The answer is:

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Example Question #4279 : Algebra Ii

Solve: $\frac{\sqrt{3x}}{x} = 6$

Possible Answers:

$\frac{1}{12}$

$\frac{2}{5}$

$\frac{1}{3}$

$\frac{1}{24}$

$\frac{1}{42}$

Correct answer:

$\frac{1}{12}$

Explanation:

Multiply by on both sides.

$\frac{\sqrt{3x}}{x} \cdot x= 6\cdot x$

Simplify both sides.

$\sqrt{3x}=6x$

Square both sides to eliminate the radical.

$(\sqrt{3x} )^2=(6x)^2$

Simplify both sides.

3x= 36x^2

Divide by 36x on both sides.

$\frac{3x}{36x}= \frac{36x^2}{36x}$

The answer is: $x=\frac{1}{12}$

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Example Question #4280 : Algebra Ii

Solve the radical: $2\sqrt[3]{3x} = \frac{2}{3}$

Possible Answers:

$\frac{1}{81}$

$\frac{3}{28}$

$\frac{64}{9}$

$\frac{1}{18}$

$\frac{1}{9}$

Correct answer:

$\frac{1}{81}$

Explanation:

Divide by two on both sides. This is similar to multiplying by one-half.

$\frac{2\sqrt[3]{3x}}{2} = \frac{2}{3}\cdot \frac{1}{2}$

Simplify both sides.

$\sqrt[3]{3x} = \frac{1}{3}$

Cube both sides to eliminate the radical.

$(\sqrt[3]{3x} )^3= (\frac{1}{3})^3$

$3x=\frac{1}{27}$

Divide by three on both sides. This is similar to multiplying by one-third on both sides.

$\frac{3x}{3}=\frac{1}{27} \cdot \frac{1}{3}$

The answer is: $\frac{1}{81}$

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Example Question #61 : Solving And Graphing Radicals

Solve the equation: $\sqrt{x-9}+31 = 35$

Possible Answers:

$\textup{There is no solution.}$

169

Correct answer:

Explanation:

Subtract 31 from both sides of the equation.

$\sqrt{x-9}+31-31 = 35-31$

$\sqrt{x-9}=4$

Square both sides.

$(\sqrt{x-9})^2=4^2$

Simplify both sides.

x-9 =16

Add nine on both sides.

x-9 +9=16+9

The answer is:

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Example Question #381 : Radicals

Solve the equation: $\sqrt[3]{3x-7}=4$

Possible Answers:

$\frac{131}{9}$

$\frac{57}{6}$

$\frac{57}{3}$

$\frac{71}{3}$

$\textup{No solution.}$

Correct answer:

$\frac{71}{3}$

Explanation:

To eliminate the radical, we will have to cube both sides.

$(\sqrt[3]{3x-7})^3=4^3$

The equation becomes:

3x-7=64

Add seven on both sides.

3x=71

Divide by three on both sides.

$\frac{3x}{3}=\frac{71}{3}$

The answer is: $\frac{71}{3}$

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Example Question #61 : Solving Radical Equations

Solve: $3\sqrt[4]{9-2x} = 1$

Possible Answers:

$\frac{364}{81}$

$\frac{1459}{162}$

$\frac{99}{2}$

$-\frac{364}{81}$

Correct answer:

$\frac{364}{81}$

Explanation:

Divide by three on both sides.

$\frac{3\sqrt[4]{9-2x} }{3}= \frac{1}{3}$

Rewrite the equation.

$\sqrt[4]{9-2x} = \frac{1}{3}$

Raise both sides to the fourth power.

$(\sqrt[4]{9-2x})^4 =( \frac{1}{3})^4$

Simplify both sides.

$9-2x = \frac{1}{81}$

Multiply by 81 on both sides to eliminate the fraction.

$81(9-2x )= \frac{1}{81} \cdot 81$

729-162x=1

Subtract 729 from both sides.

729-162x-729=1-729

-162x = -728

Divide by negative 162 on both sides.

$\frac{-162x}{-162} =\frac{ -728}{-162}$

Reduce the fractions. Double negatives will negate, turning the answer positive.

$\frac{ -728}{-162} =\frac{364\times 2}{81\times 2}$

The answer is: $\frac{364}{81}$

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Example Question #382 : Radicals

Evaluate: $3\sqrt[4]{4-2x}-4=2$

Possible Answers:

$-\frac{1}2$

Correct answer:

Explanation:

Isolate the radical. Add four on both sides.

$3\sqrt[4]{4-2x}-4+4=2+4$

$3\sqrt[4]{4-2x}=6$

Divide by three on both sides. The equation becomes:

$\sqrt[4]{4-2x}= 2$

Raise both sides by four.

$(\sqrt[4]{4-2x}) ^4= 2^4$

4-2x = 16

Subtract four on both sides.

4-2x -4= 16-4

-2x=12

Divide by negative two on both sides.

$\frac{-2x}{-2}=\frac{12}{-2}$

The answer is:

Report an Error

Example Question #383 : Radicals

Solve the equation: $\frac{\sqrt[3]{5x-2}}{6} = \frac{1}{2}$

Possible Answers:

346

$\frac{514}{5}$

$\frac{29}{5}$

$-\frac{11}{27}$

$\frac{11}{27}$

Correct answer:

$\frac{29}{5}$

Explanation:

Multiply by six on both sides of the equation.

$\frac{\sqrt[3]{5x-2}}{6} \cdot 6= \frac{1}{2} \cdot 6$

The equation becomes:

$\sqrt[3]{5x-2} =3$

Cube both sides to eliminate the radical.

$(\sqrt[3]{5x-2}) ^3=3^3$

5x-2 = 27

Add two on both sides.

5x-2 +2= 27+2

5x=29

Divide by five on both sides.

$\frac{5x}{5}=\frac{29}{5}$

The answer is: $\frac{29}{5}$

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Example Question #384 : Radicals

Solve the radical equation: $\sqrt{-6x-5}-7 = 10$

Possible Answers:

$-\frac{1}{7}$

$-\frac{49}{11}$

$\textup{No solution.}$

$-\frac{7}{3}$

Correct answer:

Explanation:

Add seven on both sides.

$\sqrt{-6x-5}-7 +7= 10+7$

$\sqrt{-6x-5} = 17$

Square both sides. This will eliminate the radical.

$(\sqrt{-6x-5})^2= 17^2$

-6x-5 = 289

Add five on both sides.

-6x-5 +5= 289+5

-6x=294

Divide both sides by negative 6 and reduce the fraction.

$\frac{-6x}{-6}=\frac{294}{-6}$

The answer is:

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Algebra II : Solving Radical Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #4277 : Algebra Ii

Example Question #4278 : Algebra Ii

Example Question #4279 : Algebra Ii

Example Question #4280 : Algebra Ii

Example Question #61 : Solving And Graphing Radicals

Example Question #381 : Radicals

Example Question #61 : Solving Radical Equations

Example Question #382 : Radicals

Example Question #383 : Radicals

Example Question #384 : Radicals

All Algebra II Resources

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