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Algebra II : Solving and Graphing Radicals

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Example Questions

Example Question #81 : Solving Radical Equations

Solve the equation: $\sqrt[3]{-6x-9} = 4$ $\displaystyle \sqrt[3]{-6x-9} = 4$

Possible Answers:

$-\frac{73}{6}$ $\displaystyle -\frac{73}{6}$

$-\frac{73}{3}$ $\displaystyle -\frac{73}{3}$

$-\frac{13}{6}$ $\displaystyle -\frac{13}{6}$

$-\frac{65}{6}$ $\displaystyle -\frac{65}{6}$

$-\frac{7}{3}$ $\displaystyle -\frac{7}{3}$

Correct answer:

$-\frac{73}{6}$ $\displaystyle -\frac{73}{6}$

Explanation:

Raise both sides by the power of three in order to eliminate the radical.

$(\sqrt[3]{-6x-9} )^3= 4^3$ $\displaystyle (\sqrt[3]{-6x-9} )^3= 4^3$

$\displaystyle -6x-9 =64$

Add nine on both sides.

$\displaystyle -6x-9 +9=64+9$

-6x =73 $\displaystyle -6x =73$

Divide by negative six on both sides.

$\frac{-6x}{-6} =\frac{73}{-6}$ $\displaystyle \frac{-6x}{-6} =\frac{73}{-6}$

The answer is: $-\frac{73}{6}$ $\displaystyle -\frac{73}{6}$

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Example Question #82 : Solving Radical Equations

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

Possible Answers:

$\frac{37}{18}$ $\displaystyle \frac{37}{18}$

$\frac{49}{24}$ $\displaystyle \frac{49}{24}$

$-\frac{35}{18}$ $\displaystyle -\frac{35}{18}$

$-\frac{47}{24}$ $\displaystyle -\frac{47}{24}$

$-\frac{49}{24}$ $\displaystyle -\frac{49}{24}$

Correct answer:

$\frac{49}{24}$ $\displaystyle \frac{49}{24}$

Explanation:

Cube both sides to eliminate the radical.

$(\sqrt[3]{3x-6} )^3= (\frac{1}{2})^3$ $\displaystyle (\sqrt[3]{3x-6} )^3= (\frac{1}{2})^3$

The equation becomes: $3x-6 = \frac{1}{8}$ $\displaystyle 3x-6 = \frac{1}{8}$

Multiply by eight on both sides to eliminate the fraction.

$8(3x-6)= \frac{1}{8}\cdot 8$ $\displaystyle 8(3x-6)= \frac{1}{8}\cdot 8$

$\displaystyle 24x-48=1$

Add 48 on both sides.

$\displaystyle 24x-48+48=1+48$

24x=49 $\displaystyle 24x=49$

Divide by 24 on both sides.

$\frac{24x}{24}=\frac{49}{24}$ $\displaystyle \frac{24x}{24}=\frac{49}{24}$

The answer is: $\frac{49}{24}$ $\displaystyle \frac{49}{24}$

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Example Question #83 : Solving Radical Equations

Solve: $\sqrt{9x}+2 = 16$ $\displaystyle \sqrt{9x}+2 = 16$

Possible Answers:

$\frac{196}{9}$ $\displaystyle \frac{196}{9}$

$\frac{9}{49}$ $\displaystyle \frac{9}{49}$

$\displaystyle 36$

$\frac{62}{3}$ $\displaystyle \frac{62}{3}$

$\frac{1}{7}$ $\displaystyle \frac{1}{7}$

Correct answer:

$\frac{196}{9}$ $\displaystyle \frac{196}{9}$

Explanation:

Subtract two from both sides.

$\sqrt{9x}+2 -2= 16-2$ $\displaystyle \sqrt{9x}+2 -2= 16-2$

$\sqrt{9x} = 14$ $\displaystyle \sqrt{9x} = 14$

Square both sides in order to eliminate the radical.

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

9x=196 $\displaystyle 9x=196$

Divide by nine on both sides.

$\frac{9x}{9}=\frac{196}{9}$ $\displaystyle \frac{9x}{9}=\frac{196}{9}$

The right side is irreducible.

The answer is: $\frac{196}{9}$ $\displaystyle \frac{196}{9}$

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Example Question #84 : Solving Radical Equations

Solve the radical: $\sqrt{-5x+3} = 3$ $\displaystyle \sqrt{-5x+3} = 3$

Possible Answers:

$-\frac{6}{5}$ $\displaystyle -\frac{6}{5}$

$\displaystyle 0$

$-\frac{1}{5}$ $\displaystyle -\frac{1}{5}$

$-\frac{45}{4}$ $\displaystyle -\frac{45}{4}$

$-\frac{5}{6}$ $\displaystyle -\frac{5}{6}$

Correct answer:

$-\frac{6}{5}$ $\displaystyle -\frac{6}{5}$

Explanation:

Square both sides in order to eliminate the radical.

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

The equation becomes:

-5x+3=9 $\displaystyle -5x+3=9$

Subtract three from both sides.

$\displaystyle -5x+3-3=9-3$

-5x=6 $\displaystyle -5x=6$

Divide by negative five on both sides.

$\frac{-5x}{-5}=\frac{6}{-5}$ $\displaystyle \frac{-5x}{-5}=\frac{6}{-5}$

The answer is: $-\frac{6}{5}$ $\displaystyle -\frac{6}{5}$

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Example Question #401 : Radicals

Solve: $\sqrt[5]{-3x}-3 = -1$ $\displaystyle \sqrt[5]{-3x}-3 = -1$

Possible Answers:

$-\frac{20}{3}$ $\displaystyle -\frac{20}{3}$

$\textup{No solution.}$ $\displaystyle \textup{No solution.}$

$-\frac{2}{3}$ $\displaystyle -\frac{2}{3}$

$-\frac{16}{3}$ $\displaystyle -\frac{16}{3}$

$-\frac{32}{3}$ $\displaystyle -\frac{32}{3}$

Correct answer:

$-\frac{32}{3}$ $\displaystyle -\frac{32}{3}$

Explanation:

Add three on both sides.

$\sqrt[5]{-3x}-3+3 = -1+3$ $\displaystyle \sqrt[5]{-3x}-3+3 = -1+3$

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

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

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

-3x =32 $\displaystyle -3x =32$

Divide by negative three on both sides.

$\frac{-3x }{-3}=\frac{32}{-3}$ $\displaystyle \frac{-3x }{-3}=\frac{32}{-3}$

The answer is: $-\frac{32}{3}$ $\displaystyle -\frac{32}{3}$

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Example Question #402 : Radicals

Solve the equation: $\sqrt5 \cdot \sqrt{2x} -10= -9$ $\displaystyle \sqrt5 \cdot \sqrt{2x} -10= -9$

Possible Answers:

$\textup{No solution.}$ $\displaystyle \textup{No solution.}$

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

$\frac{1}{10}$ $\displaystyle \frac{1}{10}$

$\frac{1}{20}$ $\displaystyle \frac{1}{20}$

$\frac{1}{5}$ $\displaystyle \frac{1}{5}$

Correct answer:

$\frac{1}{10}$ $\displaystyle \frac{1}{10}$

Explanation:

The radicals can be combined as one radical by multiplying the two inner terms together.

The equation becomes:

$\sqrt{10x} -10= -9$ $\displaystyle \sqrt{10x} -10= -9$

Add 10 on both sides.

$\sqrt{10x} -10+10= -9+10$ $\displaystyle \sqrt{10x} -10+10= -9+10$

Simplify both sides.

$\sqrt{10x} =1$ $\displaystyle \sqrt{10x} =1$

Square both sides.

10x=1 $\displaystyle 10x=1$

Divide by ten on both sides.

$\frac{10x}{10}=\frac{1}{10}$ $\displaystyle \frac{10x}{10}=\frac{1}{10}$

The answer is: $\frac{1}{10}$ $\displaystyle \frac{1}{10}$

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Example Question #403 : Radicals

Solve the radical: $\sqrt{\sqrt{2x}} = \sqrt{8}$ $\displaystyle \sqrt{\sqrt{2x}} = \sqrt{8}$

Possible Answers:

128 $\displaystyle 128$

$\displaystyle 16$

$\displaystyle 64$

$\displaystyle 32$

256 $\displaystyle 256$

Correct answer:

$\displaystyle 32$

Explanation:

Rewrite the radicals so that they are in exponential form.

$\sqrt{x} = x^\frac{1}{2}$ $\displaystyle \sqrt{x} = x^\frac{1}{2}$

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

The equation becomes:

$(2x)^{\frac{1}{4}} = 8^{\frac{1}{2}}$ $\displaystyle (2x)^{\frac{1}{4}} = 8^{\frac{1}{2}}$

Raise both sides by the power of four to eliminate the fractional exponents.

$[(2x)^{\frac{1}{4}}] ^4= [8^{\frac{1}{2}}]^4$ $\displaystyle [(2x)^{\frac{1}{4}}] ^4= [8^{\frac{1}{2}}]^4$

2x=64 $\displaystyle 2x=64$

Divide by two on both sides.

$\frac{2x}{2}=\frac{64}{2}$ $\displaystyle \frac{2x}{2}=\frac{64}{2}$

The answer is: $\displaystyle 32$

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Example Question #404 : Radicals

Solve: $4=2^{10x-8}$ $\displaystyle 4=2^{10x-8}$

Possible Answers:

$\frac{3}{4}$ $\displaystyle \frac{3}{4}$

$\displaystyle -4$

$\displaystyle 1$

$\displaystyle -1$

$-\frac{1}{2}$ $\displaystyle -\frac{1}{2}$

Correct answer:

$\displaystyle 1$

Explanation:

Evaluate by changing the base of the left term.

$2^2=2^{10x-8}$ $\displaystyle 2^2=2^{10x-8}$

Now that both bases are common, set the powers equal and solve for x.

2=10x-8 $\displaystyle 2=10x-8$

Add 8 on both sides.

$\displaystyle 2+8=10x-8+8$

10=10x $\displaystyle 10=10x$

Divide by ten on both sides.

$\frac{10}{10}=\frac{10x}{10}$ $\displaystyle \frac{10}{10}=\frac{10x}{10}$

The answer is: $\displaystyle 1$

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Example Question #405 : Radicals

Evaluate: $\sqrt[3]{2x-8} = 2$ $\displaystyle \sqrt[3]{2x-8} = 2$

Possible Answers:

$\textup{No solution.}$ $\displaystyle \textup{No solution.}$

$\displaystyle 16$

-8,8 $\displaystyle -8,8$

$\displaystyle 4$

$\displaystyle 8$

Correct answer:

$\displaystyle 8$

Explanation:

Cube both sides to eliminate the radical.

$(\sqrt[3]{2x-8}) ^3= 2^3$ $\displaystyle (\sqrt[3]{2x-8}) ^3= 2^3$

2x-8 =8 $\displaystyle 2x-8 =8$

Add eight on both sides.

$\displaystyle 2x-8+(8) =8+(8)$

2x=16 $\displaystyle 2x=16$

Divide by 2 on both sides.

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

The answer is: $\displaystyle 8$

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Example Question #406 : Radicals

Solve: $2\sqrt[3]{2-5x} = 10$ $\displaystyle 2\sqrt[3]{2-5x} = 10$

Possible Answers:

$\frac{13}{5}$ $\displaystyle \frac{13}{5}$

$-\frac{123}{5}$ $\displaystyle -\frac{123}{5}$

$\frac{123}{5}$ $\displaystyle \frac{123}{5}$

$\frac{127}{5}$ $\displaystyle \frac{127}{5}$

$-\frac{13}{5}$ $\displaystyle -\frac{13}{5}$

Correct answer:

$-\frac{123}{5}$ $\displaystyle -\frac{123}{5}$

Explanation:

Solve by first dividing both sides by two.

$\frac{2\sqrt[3]{2-5x} }{2}= \frac{10}{2}$ $\displaystyle \frac{2\sqrt[3]{2-5x} }{2}= \frac{10}{2}$

Simplify both sides.

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

Cube both sides.

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

$\displaystyle 2-5x =125$

Subtract two from both sides.

$\displaystyle 2-5x -(2)=125-(2)$

-5x=123 $\displaystyle -5x=123$

Divide by negative five on both sides.

$\frac{-5x}{-5}=\frac{123}{-5}$ $\displaystyle \frac{-5x}{-5}=\frac{123}{-5}$

The answer is: $-\frac{123}{5}$ $\displaystyle -\frac{123}{5}$

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Algebra II : Solving and Graphing Radicals

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #81 : Solving Radical Equations

Example Question #82 : Solving Radical Equations

Example Question #83 : Solving Radical Equations

Example Question #84 : Solving Radical Equations

Example Question #401 : Radicals

Example Question #402 : Radicals

Example Question #403 : Radicals

Example Question #404 : Radicals

Example Question #405 : Radicals

Example Question #406 : Radicals

All Algebra II Resources

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