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Algebra II : Radicals

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Example Questions

Example Question #411 : Radicals

Solve the equation: $\sqrt{3x}-6= 11$

Possible Answers:

$\frac{289}{3}$

$\frac{23}{2}$

$\frac{34}{3}$

Correct answer:

$\frac{289}{3}$

Explanation:

Add six on both sides.

$\sqrt{3x}-6+6= 11+6$

$\sqrt{3x} = 17$

Square both sides to eliminate the radical.

$(\sqrt{3x}) ^2= 17^2$

3x=289

Divide both sides by three.

$\frac{3x}{3}=\frac{289}{3}$

The answer is: $\frac{289}{3}$

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Example Question #412 : Radicals

Solve the radical: $\sqrt{2-3x} = 40$

Possible Answers:

$-\frac{1598}{3}$

534

$\textup{No solution.}$

$-\frac{82}{3}$

Correct answer:

$-\frac{1598}{3}$

Explanation:

Square both sides of the equation to eliminate the radical.

$(\sqrt{2-3x})^2 = 40^2$

The equation becomes:

2-3x=1600

Subtract two from both sides.

2-3x-2=1600-2

-3x= 1598

Divide by negative three on both sides.

$\frac{-3x}{-3}=\frac{ 1598}{-3}$

The answer is: $-\frac{1598}{3}$

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Example Question #413 : Radicals

Evaluate: $2\sqrt[3]{-2x} = \frac{1}{2}$

Possible Answers:

$-\frac{1}{4}$

$-\frac{1}{2}$

$-\frac{1}{64}$

$-\frac{1}{128}$

$\textup{No solution.}$

Correct answer:

$-\frac{1}{128}$

Explanation:

Multiply by half on both sides.

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

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

Cube both sides to eliminate the radical.

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

$-2x = \frac{1}{64}$

Divide both sides by negative two. This is the same as multiplying by negative half on the right side.

$\frac{-2x}{-2} = \frac{1}{64}\cdot -\frac{1}{2}$

Reduce both fractions.

The answer is: $-\frac{1}{128}$

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Example Question #412 : Radicals

Solve the radical equation: $\frac{x}{\sqrt{x-9}} = 6$

Possible Answers:

$\textup{No solution.}$

Correct answer:

Explanation:

Multiply both sides by the denominator to eliminate the fraction on the left side.

$\frac{x}{\sqrt{x-9}}\cdot \sqrt{x-9}= 6\cdot \sqrt{x-9}$

The equation becomes:

$x= 6\sqrt{x-9}$

Square both sides to eliminate the radical.

$x^2=( 6\sqrt{x-9})^2$

x^2 = 36(x-9)

Distribute the 36 into the binomial.

x^2 = 36x-324

Subtract 36x and add 324 on both sides of the equation.

x^2 -[36x]+(324)= 36x-324-[36x]+(324)

The equation becomes a parabola, which can be factored.

x^2-36x+324 = 0

(x-18)^2 = 0

The value of x=18 . Check this value to see if substitution will satisfy the domain in the original equation.

$\frac{18}{\sqrt{18-9}}$ is valid since there is no negative in the radical, and there is not a zero denominator.

The answer is:

Report an Error

Example Question #99 : Solving Radical Equations

Solve the equation: $\sqrt{-9x} +5=9$

Possible Answers:

$-\frac{196}{9}$

$\textup{No solution.}$

$\frac{4}{3}$

$-\frac{16}{9}$

Correct answer:

$-\frac{16}{9}$

Explanation:

Subtract both sides by five.

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

$\sqrt{-9x} =4$

Square both sides.

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

-9x = 16

Divide by negative nine on both sides.

$\frac{-9x}{-9} = \frac{16}{-9}$

The answer is: $-\frac{16}{9}$

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Example Question #1652 : Mathematical Relationships And Basic Graphs

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

Possible Answers:

$\frac{1}{8}$

$\frac{3}{4}$

$\frac{1}{4}$

$\textup{No solution.}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{4}$

Explanation:

Add the radical on both sides.

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

The equation becomes:

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

Subtract 1 on both sides.

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

$1=\sqrt[3]{4x}$

Cube both sides to eliminate the radical.

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

1=4x

Divide by four on both sides.

$\frac{1}{4}=\frac{4x}{4}$

The answer is: $\frac{1}{4}$

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Example Question #101 : Solving Radical Equations

Solve the radical equation: $\frac{3}{\sqrt[4]{4x}}=2$

Possible Answers:

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

$\frac{9}{8}$

$\frac{9}{64}$

$\frac{81}{8}$

$\frac{81}{64}$

Correct answer:

$\frac{81}{64}$

Explanation:

Multiply the radical denominator on both sides.

$\frac{3}{\sqrt[4]{4x}}\cdot \sqrt[4]{4x}=2\cdot \sqrt[4]{4x}$

The equation becomes:

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

Divide by two on both sides.

$\frac{3}{2}=\frac{2\cdot \sqrt[4]{4x}}{2}$

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

Raise both sides by the power of four to eliminate the radical.

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

$\frac{81}{16}= 4x$

Divide by four on both sides. This is the same as multiplying one-fourth on both sides.

$\frac{81}{16}\cdot \frac{1}{4}= 4x\cdot \frac{1}{4}$

The answer is: $\frac{81}{64}$

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Example Question #101 : Solving Radical Equations

Solve: $\frac{9\sqrt2}{\sqrt{2x}} = \sqrt2$

Possible Answers:

$\frac{9}{2}$

$\frac{3}{4}$

$\frac{81}{2}$

$\frac{2}{3}$

$\frac{81}{4}$

Correct answer:

$\frac{81}{2}$

Explanation:

Multiply the denominator on both sides to eliminate the denominator.

$\frac{9\sqrt2}{\sqrt{2x}} \cdot\sqrt{2x}= \sqrt2 \cdot\sqrt{2x}$

$9\sqrt2 = \sqrt{4x}$

Square both sides.

$(9\sqrt2)^2 =( \sqrt{4x})^2$

$81\cdot 2 = 4x$

Divide by four on both sides.

$\frac{81\cdot 2}{4} =\frac{ 4x}{4}$

The answer is: $\frac{81}{2}$

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Example Question #4314 : Algebra Ii

Solve the equation: $\sqrt{5x-9} = 5$

Possible Answers:

$\frac{19}{25}$

$\frac{14}{5}$

$\frac{196}{5}$

$\frac{9}{5}$

$\frac{34}{5}$

Correct answer:

$\frac{34}{5}$

Explanation:

Square both sides of the radical to eliminate the square root.

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

The equation becomes:

5x-9 = 25

Add 9 on both sides.

5x-9 +9= 25+9

5x=34

Divide by 5 on both sides.

$\frac{5x}{5}=\frac{34}{5}$

The answer is: $\frac{34}{5}$

Report an Error

Example Question #4315 : Algebra Ii

Evaluate: $\frac{5\sqrt x}{\sqrt{9x}} = x$

Possible Answers:

$-\frac{5}{3}$

$\frac{5}{3}$

$\frac{\sqrt{5}}{3}$

$\frac{3\sqrt5}{5}$

$\frac{1}{3}$

Correct answer:

$\frac{5}{3}$

Explanation:

Divide terms of the radicals on the left side.

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

$x=5\sqrt{\frac{1}{9}} = 5\cdot \frac{1}{3} = \frac{5}{3}$

The answer is: $\frac{5}{3}$

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Algebra II : Radicals

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #411 : Radicals

Example Question #412 : Radicals

Example Question #413 : Radicals

Example Question #412 : Radicals

Example Question #99 : Solving Radical Equations

Example Question #1652 : Mathematical Relationships And Basic Graphs

Example Question #101 : Solving Radical Equations

Example Question #101 : Solving Radical Equations

Example Question #4314 : Algebra Ii

Example Question #4315 : Algebra Ii

All Algebra II Resources

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