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Algebra II : Solving Radical Equations

Study concepts, example questions & explanations for Algebra II

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10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #31 : Solving Radical Equations

Solve the equation: $\sqrt{6-x}+2 = 16$

Possible Answers:

$\textup{There is no solution.}$

400

Correct answer:

Explanation:

Subtract two from both sides.

$\sqrt{6-x}+2 -(2)= 16-(2)$

Simplify the equation.

$\sqrt{6-x} = 14$

Square both sides.

$(\sqrt{6-x})^2 =( 14)^2$

6-x=196

Add on both sides.

6-x+x=196+x

6=196+x

Subtract 196 on both sides.

6-196=196+x-196

The answer is:

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Example Question #32 : Solving Radical Equations

Solve: $\sqrt{9x+2}-4=10$

Possible Answers:

$\textup{There is no solution.}$

$\frac{98}{9}$

$\frac{172}{9}$

$\frac{194}{9}$

$\frac{26}{9}$

Correct answer:

$\frac{194}{9}$

Explanation:

Add four on both sides to isolate the radical.

$\sqrt{9x+2}-4+4=10+4$

Simplify both sides.

$\sqrt{9x+2}= 14$

Square both sides.

$(\sqrt{9x+2})^2= (14)^2$

9x+2=196

Subtract two from both sides.

9x=194

Divide by nine on both sides.

$\frac{9x}{9}=\frac{194}{9}$

The answer is: $\frac{194}{9}$

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Example Question #33 : Solving Radical Equations

Solve the equation: $\sqrt{2x-3}-19=6$

Possible Answers:

314

$\textup{The answer is not given.}$

376

Correct answer:

314

Explanation:

Add 19 on both sides.

$\sqrt{2x-3}-19+(19)=6+(19)$

Simplify both sides.

$\sqrt{2x-3} = 25$

Square both sides.

$(\sqrt{2x-3})^2= 25^{2}$

2x-3=625

Add three on both sides.

2x-3+3=625+3

2x=628

Divide by two on both sides.

$\frac{2x}{2}=\frac{628}{2}$

Reduce the fractions.

The answer is: 314

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Example Question #34 : Solving Radical Equations

Solve the equation: $\sqrt{-3x-9} =3$

Possible Answers:

$-\frac{1}{6}$

$\textup{No solution.}$

Correct answer:

Explanation:

Square both sides of the equation.

$(\sqrt{-3x-9}) ^2=3^2$

Simplify both sides.

-3x-9 = 9

Add nine on both sides.

-3x-9 +9= 9+9

-3x=18

Divide both sides by negative three.

$\frac{-3x}{-3}=\frac{18}{-3}$

Simplify both fractions.

x=-6

The answer is:

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Example Question #1581 : Mathematical Relationships And Basic Graphs

Solve the equation: $\sqrt{3-3x}+6 =10$

Possible Answers:

$-\frac{7}{3}$

$-\frac{13}{3}$

$-\frac{7}{3}$

$-\frac{19}{3}$

$\textup{No solution.}$

Correct answer:

$-\frac{13}{3}$

Explanation:

Subtract six from both sides.

$\sqrt{3-3x}+6-6 =10-6$

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

Square both sides.

$(\sqrt{3-3x})^2=4^2$

3-3x=16

Add on both sides.

3-3x+3x=16+3x

3=16+3x

Subtract 16 from both sides.

3-16=16+3x-16

-13 = 3x

Divide by three on both sides.

$\frac{-13 }{3}=\frac{ 3x}{3}$

The answer is: $-\frac{13}{3}$

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Example Question #31 : Solving And Graphing Radicals

Solve the equation: $-\sqrt{-9x}-3 =- 8$

Possible Answers:

$-\frac{4}{3}$

$\textup{There is no solution.}$

$-\frac{31}{9}$

$-\frac{5}{9}$

$-\frac{25}{9}$

Correct answer:

$-\frac{25}{9}$

Explanation:

Add three on both sides.

$-\sqrt{-9x}-3 +3=- 8+3$

$-\sqrt{-9x} = -5$

Divide by negative one to isolate the radical.

$\frac{-\sqrt{-9x}}{-1} = \frac{-5}{-1}$

The equation becomes: $\sqrt{-9x} =5$

Square both sides.

$(\sqrt{-9x})^2 =5^2$

-9x=25

Divide by negative nine on both sides.

$\frac{-9x}{-9}=\frac{25}{-9}$

The answer is: $-\frac{25}{9}$

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Example Question #33 : Solving Radical Equations

Solve the equation: $\sqrt{5x+3}+16= 25$

Possible Answers:

412

$\frac{1678}{5}$

$\frac{78}{5}$

$\frac{84}{5}$

Correct answer:

$\frac{78}{5}$

Explanation:

Solve by first subtracting 16 on both sides.

$\sqrt{5x+3}+16-16= 25-16$

The equation becomes:

$\sqrt{5x+3} =9$

Square both sides.

$(\sqrt{5x+3})^2 =9^2$

5x+3=81

Subtract three on both sides.

5x+3-3=81-3

5x=78

Divide by five on both sides.

$\frac{5x}{5}=\frac{78}{5}$

The answer is: $\frac{78}{5}$

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Example Question #1592 : Mathematical Relationships And Basic Graphs

Solve the equation: $\sqrt[3]{3x}+6=8$

Possible Answers:

$\frac{3}{2}$

$\frac{8}{3}$

Correct answer:

$\frac{8}{3}$

Explanation:

Subtract six from both sides.

$\sqrt[3]{3x}+6-6=8-6$

Simplify both sides.

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

Cube both sides to eliminate the cube root.

3x= 2^3

3x=8

Divide by three on both sides.

$\frac{3x}{3}=\frac{8}{3}$

The answer is: $\frac{8}{3}$

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Example Question #352 : Radicals

Solve the equation: $\sqrt[4]{3(x+1)}+16=18$

Possible Answers:

$\frac{26}{3}$

$\frac{26}{5}$

$\textup{There is no solution.}$

$\frac{13}{3}$

Correct answer:

$\frac{13}{3}$

Explanation:

Simplify by first subtracting 16 on both sides.

$\sqrt[4]{3(x+1)}+16-16=18-16$

$\sqrt[4]{3(x+1)}=2$

Raise both sides by the fourth power.

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

3(x+1)=16

Expand the binomial.

3x+3=16

Subtract three from both sides.

3x+3-3=16-3

3x=13

Divide by three on both sides.

$\frac{3x}{3}=\frac{13}{3}$

The answer is: $\frac{13}{3}$

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Example Question #32 : Solving And Graphing Radicals

Solve the equation: $\sqrt[3]{9-x} = 3$

Possible Answers:

$\textup{There is no solution.}$

Correct answer:

Explanation:

To eliminate the radical, we will need to cube both sides of the equation.

$(\sqrt[3]{9-x})^3 = 3^3$

The equation becomes:

9-x = 27

Add on both sides.

9-x+(x) = 27+(x)

9=27+x

Subtract 27 from both sides.

9-27=27+x-27

x=-18

This answer is valid after re-substituting back to the original equation.

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 #31 : Solving Radical Equations

Example Question #32 : Solving Radical Equations

Example Question #33 : Solving Radical Equations

Example Question #34 : Solving Radical Equations

Example Question #1581 : Mathematical Relationships And Basic Graphs

Example Question #31 : Solving And Graphing Radicals

Example Question #33 : Solving Radical Equations

Example Question #1592 : Mathematical Relationships And Basic Graphs

Example Question #352 : Radicals

Example Question #32 : Solving And Graphing Radicals

All Algebra II Resources

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