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Algebra II : Solving Radical Equations

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Example Question #331 : Radicals

Solve and simplify:

$2\sqrt{x^3}=8$

Possible Answers:

$x=\sqrt{2}$

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

$x=\sqrt[3]{16}$

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

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

Correct answer:

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

Explanation:

To solve for x, first we must isolate the radical on one side:

$\sqrt{x^3}=4$

Next, square both sides to eliminate the radical:

x^3=16

Now, take the cube root of each side to find x:

$x=\sqrt[3]{16}$

Finally, factor the term inside the cube root and see if any cubes can be pulled out of the radical:

$x=\sqrt[3]{16}=\sqrt[3]{8\cdot 2}=2\sqrt[3]{2}$

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Example Question #4231 : Algebra Ii

Solve: $\sqrt{2x} + 14 = 19$

Possible Answers:

$\pm\frac{25}{2}$

$\textup{There is no solution.}$

$\frac{25}{8}$

$-\frac{25}{2}$

$\frac{25}{2}$

Correct answer:

$\frac{25}{2}$

Explanation:

Subtract 14 on both sides.

$\sqrt{2x} + 14 -14= 19-14$

Simplify both sides.

$\sqrt{2x} = 5$

To eliminate the radical, square both sides.

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

Simplify both sides.

2x=25

Divide by two on both sides.

$\frac{2x}{2}=\frac{25}{2}$

The answer is: $\frac{25}{2}$

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Example Question #23 : Solving And Graphing Radicals

Solve the equation: $\sqrt{2x}+1 = \sqrt{x+3}$

Possible Answers:

$4\sqrt{2}-3$

$2\sqrt2 -8$

$6-4\sqrt{2}$

$\textup{There is no solution.}$

$3-4\sqrt{2}$

Correct answer:

$6-4\sqrt{2}$

Explanation:

Subtract $\sqrt{2x}$ from both sides to group the radicals.

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

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

Square both sides.

$1^2= (\sqrt{x+3}-\sqrt{2x}) (\sqrt{x+3}-\sqrt{2x})$

Use the FOIL method to simplify the right side.

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

$1=(x+3)-2\sqrt{2x(x+3)}+2x$

Combine like-terms.

$1=3x-2\sqrt{2x(x+3)}+3$

Subtract one from both sides, and add $2\sqrt{2x(x+3)}$ on both sides.

The equation becomes: $2\sqrt{2x(x+3)}= 3x+2$

Divide by two on both sides and distribute the terms inside the radical.

$\sqrt{2x^2+6x} = \frac{3}{2}x+1$

Square both sides.

$(\sqrt{2x^2+6x} )^2=( \frac{3}{2}x+1)^2$

Simplify the right side by FOIL method.

$2x^2+6x = \frac{9}{4}x^2+3x+1$

Subtract 2x^2 on both sides. This is the same as subtracting $\frac{8}{4}x^2$ on both sides.

$2x^2+6x -2x^2 = (\frac{9}{4}x^2-\frac{8}{4}x^2)+3x+1$

$6x = \frac{1}{4}x^2+3x+1$

Subtract on both sides. The equation will become:

$0= \frac{1}{4}x^2-3x+1$

Multiply by four on both sides to eliminate the fractional denominator.

$(0)(4)= 4(\frac{1}{4}x^2-3x+1)$

0 = x^2-12x+4

Use the quadratic equation to solve for the roots.

$x = \frac{-(-12)\pm \sqrt{(-12)^2-4(1)(4)}}{2(1)}$

Simplify the radical and fraction.

$x = \frac{12\pm \sqrt{144-16}}{2} = \frac{12 \pm \sqrt{128}}{2} = \frac{12 \pm \sqrt{64} \cdot \sqrt{2}}{2}$

$x=\frac{12\pm 8\sqrt2}{2} = 6\pm 4\sqrt2$

Substitute the values of $6+4\sqrt{2}$ and $6-4\sqrt{2}$ back into the original equation, and only $6-4\sqrt{2}$ will satisfy both sides of the equation.

The answer is: $6-4\sqrt{2}$

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Example Question #24 : Solving And Graphing Radicals

Solve, and ensure there are no radicals in the denominator

$\frac{\sqrt{356}}{\sqrt{944}}$

Possible Answers:

$\frac{\sqrt{6571}}{116}$

$\frac{\sqrt{5251}}{236}$

$\frac{\sqrt{5251}}{59}$

$\frac{\sqrt{5251}}{118}$

None of these

Correct answer:

$\frac{\sqrt{5251}}{118}$

Explanation:

$\frac{\sqrt{356}}{\sqrt{944}}$

$\frac{\sqrt{4*89}}{\sqrt{4*4*59}}$

$\frac{\sqrt{89}}{2\sqrt{59}}$

$\frac{\sqrt{89}}{2\sqrt{59}}*\frac{\sqrt{59}}{\sqrt{59}}$

$\frac{\sqrt{89}\sqrt{59}}{2*59}$

$\frac{\sqrt{5251}}{118}$

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Example Question #21 : Solving Radical Equations

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

Possible Answers:

$-2\pm i\sqrt{5}$

$-5\pm i\sqrt{2}$

$-2 i\sqrt{5}$

$-5\pm i\sqrt{2}$

$4+\sqrt{7}$

Correct answer:

$-2\pm i\sqrt{5}$

Explanation:

Add three on both sides.

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

$\sqrt{2x}=x+3$

Square both sides.

2x=(x+3)^2

Use the FOIL method to expand the right side.

2x=(x)(x)+(x)(3)+3(x)+3(3)

2x=x^2+6x+9

Subtract from both sides. The equation becomes:

0=x^2+4x+9

Use the quadratic equation to determine the roots.

$x=\frac{-4\pm \sqrt{(4^2)-4(1)(9)}}{2(1)}= \frac{-4\pm \sqrt{16-36}}{2}$

Simplify the radical.

$= \frac{-4\pm \sqrt{-20}}{2} = \frac{-4\pm( \sqrt{-1}\times \sqrt{2}\times \sqrt{2} \times \sqrt{5})}{2}$

$=\frac{-4\pm 2i\sqrt5}{2}$

Reduce the fraction.

$=-2\pm i\sqrt{5}$

The answers are: $-2\pm i\sqrt{5}$

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Example Question #4242 : Algebra Ii

Solve: $\sqrt{3x}-5 =8$

Possible Answers:

$\frac{1}{27}$

$\frac{1}{9}$

$\frac{169}{3}$

Correct answer:

$\frac{169}{3}$

Explanation:

Add five on both sides.

$\sqrt{3x}-5 +5=8+5$

Simplify both sides of the equation.

$\sqrt{3x}=13$

Square both sides.

$(\sqrt{3x})^2=(13)^2$

3x=169

Divide both sides by three.

$\frac{3x}{3}=\frac{169}{3}$

$x=\frac{169}{3}$

The answer is: $\frac{169}{3}$

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Example Question #4243 : Algebra Ii

Solve: $\sqrt{3x+2}+3= 10$

Possible Answers:

$\frac{47}{3}$

$\frac{194}{7}$

$\frac{9}{4}$

$\frac{194}{3}$

$\frac{37}{3}$

Correct answer:

$\frac{47}{3}$

Explanation:

To isolate the x-variable, first subtract three from both sides.

$\sqrt{3x+2}+3-3= 10-3$

$\sqrt{3x+2}=7$

Square both sides.

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

The equation becomes:

3x+2=49

Subtract two from both sides.

3x+2-2=49-2

3x=47

Divide by three on both sides.

$x=\frac{47}{3}$

The answer is: $\frac{47}{3}$

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Example Question #4244 : Algebra Ii

Solve: $\sqrt{3x-4}-8 = -10$

Possible Answers:

$\frac{22}{3}$

$\textup{No solution.}$

$\frac{3}{8}$

$\frac{8}{3}$

$\pm \frac{8}{3}$

Correct answer:

$\textup{No solution.}$

Explanation:

Add eight on both sides.

$\sqrt{3x-4}-8 +(8)= -10+(8)$

$\sqrt{3x-4}=-2$

Notice that the range of the parent function $y=\sqrt{x}$ must be greater than zero. There will be no values of square root that will equal to negative two.

The answer is: $\textup{No solution.}$

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Example Question #4245 : Algebra Ii

Solve the equation: $\sqrt{x+2} =\sqrt{ 2(x-1)}$

Possible Answers:

$\textup{There is no solution.}$

$2\sqrt2$

$\sqrt2$

Correct answer:

Explanation:

In order to eliminate the radicals, simply square both sides.

$(\sqrt{x+2} )^2=(\sqrt{ 2(x-1)})^2$

The equation becomes:

x+2= 2(x-1)

Simplify the right side by distribution.

x+2= 2x-2

Subtract from both sides. The equation becomes:

2= x-2

Add 2 on both sides and simplify both sides.

2+2= x-2+2

The answer is: x=4

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Example Question #4246 : Algebra Ii

Solve the equation: $2\sqrt{2x-3} = 4$

Possible Answers:

$\frac{7}{2}$

$\frac{7}{3}$

$\textup{There is no solution.}$

$\frac{3}{2}$

Correct answer:

$\frac{7}{2}$

Explanation:

Divide by two on both sides.

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

The equation becomes: $\sqrt{2x-3} = 2$

Square both sides.

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

2x-3=4

Add three on both sides.

2x=7

Divide by two on both sides.

$\frac{2x}{2}=\frac{7}{2}$

The answer is: $\frac{7}{2}$

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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 #331 : Radicals

Example Question #4231 : Algebra Ii

Example Question #23 : Solving And Graphing Radicals

Example Question #24 : Solving And Graphing Radicals

Example Question #21 : Solving Radical Equations

Example Question #4242 : Algebra Ii

Example Question #4243 : Algebra Ii

Example Question #4244 : Algebra Ii

Example Question #4245 : Algebra Ii

Example Question #4246 : Algebra Ii

All Algebra II Resources

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