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Algebra II : Solving Radical Equations

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Example Questions

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 #411 : Radicals

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

Possible Answers:

$\frac{3}{4}$

$\frac{2}{3}$

$\frac{81}{4}$

$\frac{9}{2}$

$\frac{81}{2}$

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: $\frac{34}{5}$

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Example Question #412 : Radicals

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

Possible Answers:

$\frac{1}{3}$

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

$\frac{5}{3}$

$-\frac{5}{3}$

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

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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Example Question #105 : Solving Radical Equations

Solve the equation: $\frac{\sqrt{-3-7x} }{-3}= -9$

Possible Answers:

$\textup{No solution.}$

$-\frac{726}{7}$

$-\frac{732}{7}$

$\frac{732}{7}$

$\frac{12}{7}$

Correct answer:

$-\frac{732}{7}$

Explanation:

Multiply by negative three on both sides.

$\frac{\sqrt{-3-7x} }{-3} \cdot -3= -9 \cdot -3$

$\sqrt{-3-7x} = 27$

Square both sides.

$(\sqrt{-3-7x}) ^2= (27)^2$

-3-7x = 729

Add three on both sides.

-3-7x +3= 729+3

-7x=732

Divide by negative seven on both sides.

$\frac{-7x}{-7}=\frac{732}{-7}$

The answer is: $-\frac{732}{7}$

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Example Question #106 : Solving Radical Equations

Solve the equation: $\frac{5}{\sqrt{5x}} = 10$

Possible Answers:

$\frac{1}{200}$

$\frac{\sqrt{5}}{10}$

$\frac{1}{20}$

$\frac{5}{4}$

$\frac{1}{10}$

Correct answer:

$\frac{1}{20}$

Explanation:

Multiply the denominator on both sides to eliminate the fraction.

$\frac{5}{\sqrt{5x}} \cdot \sqrt{5x}= 10 \cdot \sqrt{5x}$

The equation becomes:

$5=10\sqrt{5x}$

Square both sides.

$5^2=(10\sqrt{5x})^2$

25= 100(5x)

25 =500x

Divide by 500 on both sides.

$\frac{25}{500} =\frac{500x}{500}$

Reduce both fractions.

The answer is: $\frac{1}{20}$

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Example Question #101 : Solving Radical Equations

Solve: $\sqrt{x}-5 = \frac{1}{5}$

Possible Answers:

$\frac{24}{5}$

$\frac{2}{25}$

$\frac{676}{25}$

$\frac{50}{3}$

$\frac{1}{25}$

Correct answer:

$\frac{676}{25}$

Explanation:

Add 5 on both sides. This is the same as adding $\frac{25}{5}$ on both sides.

$\sqrt{x}-5 +5= \frac{1}{5}+\frac{25}{5}$

$\sqrt{x} = \frac{26}{5}$

To eliminate the radical, we must square both sides of the equation.

$(\sqrt{x}) ^2=( \frac{26}{5})^2$

The answer is: $\frac{676}{25}$

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Example Question #421 : Radicals

Solve: $\frac{14}{\sqrt[4]{2x}+3} =4$

Possible Answers:

$\frac{1}{16}$

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

$\frac{1}{32}$

$\frac{7}{8}$

Correct answer:

$\frac{1}{32}$

Explanation:

Multiply the denominator on both sides to eliminate the denominator.

$\frac{14}{\sqrt[4]{2x}+3} \cdot (\sqrt[4]{2x}+3)=4\cdot (\sqrt[4]{2x}+3)$

Simplify both sides.

$14 = 4\sqrt[4]{2x}+12$

Subtract 12 from both sides.

$14-12 = 4\sqrt[4]{2x}+12-12$

The equation becomes:

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

Divide by 4 on both sides.

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

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

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

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

$\frac{1}{16} = 2x$

Multiply by one half on both sides to isolate .

$\frac{1}{16}\cdot \frac{1}{2} = 2x\cdot \frac{1}{2}$

The answer is: $\frac{1}{32}$

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Example Question #101 : Solving Radical Equations

Solve: $5-3\sqrt{9x} = -2$

Possible Answers:

$-\frac{1}{9}$

$\frac{7}{9}$

$\textup{No solution.}$

$\frac{49}{81}$

$\frac{49}{3}$

Correct answer:

$\frac{49}{81}$

Explanation:

Subtract five from both sides of the equation.

$5-3\sqrt{9x} -5= -2-5$

$-3\sqrt{9x} = -7$

Divide by negative three on both sides.

$\frac{-3\sqrt{9x} }{-3}=\frac{ -7}{-3}$

$\sqrt{9x} = \frac{7}{3}$

Square both sides.

$(\sqrt{9x}) ^2=( \frac{7}{3})^2$

Simplify both sides.

$9x =\frac{49}{9}$

Multiply by one-ninth on both sides.

$9x \cdot \frac{1}{9}=\frac{49}{9}\cdot \frac{1}{9}$

The answer is: $\frac{49}{81}$

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Example Question #102 : Solving Radical Equations

Solve the equation: $\frac{\sqrt{2x}}{\sqrt5} = \frac{\sqrt6}{\sqrt x}$

Possible Answers:

$\frac{\sqrt{22}}{2}$

$\pm 2\sqrt{15}$

$\pm 15$

$\pm\frac{\sqrt{22}}{2}$

Correct answer:

Explanation:

Cross multiply both sides.

$\sqrt{2x}(\sqrt x) =(\sqrt5)(\sqrt6)$

$\sqrt{2x^2} = \sqrt{30}$

Square both sides in order to cancel the radicals.

$(\sqrt{2x^2}) ^2= (\sqrt{30})^2$

2x^2 = 30

Divide by 2 on both sides.

$\frac{2x^2}{2} = \frac{30}{2}$

x^2=15

Square root both sides.

$\sqrt{x^2} = \sqrt{15}$

$x=\pm 15$

Only x=15 will satisfy 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 #101 : Solving Radical Equations

Example Question #411 : Radicals

Example Question #4314 : Algebra Ii

Example Question #412 : Radicals

Example Question #105 : Solving Radical Equations

Example Question #106 : Solving Radical Equations

Example Question #101 : Solving Radical Equations

Example Question #421 : Radicals

Example Question #101 : Solving Radical Equations

Example Question #102 : Solving Radical Equations

All Algebra II Resources

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