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Algebra II : Solving Radical Equations

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Example Question #1 : Solving Radical Equations

Solve for :

$x + 3 \sqrt{x} = 28$

Possible Answers:

$x = 49 \textup{ or }x = 16$

None of the other responses is correct.

$x = - \sqrt{7} \textup{ or }x = 2$

$x = i \sqrt{7} \textup{ or }x = 2$

$x = -49 \textup{ or }x = 16$

Correct answer:

None of the other responses is correct.

Explanation:

One way to solve this equation is to substitute for $\sqrt{x}$ and, subsequently, $u^{2}$ for :

$x + 3 \sqrt{x} = 28$

$u^{2} + 3 u = 28$

$u^{2} + 3 u - 28 = 0$

Solve the resulting quadratic equation by factoring the expression:

(u+7)(u-4) = 0

Set each linear binomial to sero and solve:

u + 7 = 0

u = -7

u - 4 = 0

u = 4

Substitute back:

u = -7

$\sqrt{x}= -7$ - this is not possible.

u = 4

$\sqrt{x}= 4$

x = 16 - this is the only solution.

None of the responses state that x = 16 is the only solution.

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Example Question #2 : Solving Radical Equations

Solve the following radical equation.

$\frac{10\sqrt{8}}{\sqrt{16}} + 3\sqrt{2}$

Possible Answers:

$13\sqrt{2}$

$8\sqrt{2}$

$5\sqrt{8} + 3\sqrt{2}$

$\frac{10}{\sqrt{2} } +3\sqrt{2}$

Correct answer:

$8\sqrt{2}$

Explanation:

We can simplify the fraction:

$\frac{10\sqrt{8}}{\sqrt{16}} = \frac{10\sqrt{4\times 2}}{4}=\frac{10(2\sqrt{2})}{4} =\frac{20\sqrt{2}}{4}=5\sqrt{2}$

Plugging this into the equation leaves us with:

$5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}$

Note: Because they are like terms, we can add them.

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Example Question #3 : Solving Radical Equations

Solve the following radical equation.

$\sqrt{x-6} =2$

Possible Answers:

$x=6 +\sqrt{2}$

x=10

x=-10

x=2

Correct answer:

x=10

Explanation:

In order to solve this equation, we need to know that

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

How? Because of these two facts:

$\sqrt{x-6} = (x-6)^{\frac{1}{2}}$
Power rule of exponents: when we raise a power to a power, we need to mulitply the exponents:

$\frac{1}{2} \times2=1$

With this in mind, we can solve the equation:

$\sqrt{x-6} =2$

In order to eliminate the radical, we have to square it. What we do on one side, we must do on the other.

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

x-6=4

x=10

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Example Question #4 : Solving Radical Equations

Solve the following radical equation.

$\sqrt{\frac{x}{2}}=5$

Possible Answers:

x=50

$x=2\sqrt{5}$

$x=\sqrt{10}$

x=10

Correct answer:

x=50

Explanation:

In order to solve this equation, we need to know that

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

Note: This is due to the power rule of exponents.

With this in mind, we can solve the equation:

$\sqrt{\frac{x}{2}} =5$

In order to get rid of the radical we square it. Remember what we do on one side, we must do on the other.

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

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

$\frac{2}{1} \bigg(\frac{x}{2}\bigg)=25\bigg(\frac{2}{1}\bigg)$

x=50

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Example Question #5 : Solving Radical Equations

Solve for x: $\sqrt{2x+1} = 11$

Possible Answers:

$\frac{\sqrt{11} + 1 }{2}$

240

$\sqrt{5}$

Correct answer:

Explanation:

To solve, perform inverse opperations, keeping in mind order of opperations:

$\sqrt{2x+1} = 11$ first, square both sides

2x + 1 = 121 subtract 1

2x = 120 divide by 2

x = 60

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Example Question #6 : Solving Radical Equations

Solve for x: $(\sqrt{x} + 19 )^2 = 10,000$

Possible Answers:

507.3158

6,561

1,899.5

9,999.999

Correct answer:

6,561

Explanation:

To solve, perform inverse opperations, keeping in mind order of opperations:

$(\sqrt x + 19 ) ^ 2 = 10,000$ take the square root of both sides

$\sqrt x + 19 = 100$ subtract 19 from both sides

$\sqrt{x } = 81$ square both sides

x = 6,561

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Example Question #7 : Solving Radical Equations

Solve for x: $5\sqrt{x - 12 } = 20$

Possible Answers:

40.96

Correct answer:

Explanation:

To solve, use inverse opperations keeping in mind order of opperations:

$5\sqrt {x-12} = 20$ divide both sides by 5

$\sqrt{x-12 } = 4$ square both sides

x - 12 = 16 add 12 to both sides

x = 28

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Example Question #1 : Solving And Graphing Radicals

Solve for .

$\sqrt{x}=4$

Possible Answers:

$\pm2$

Correct answer:

Explanation:

$\sqrt{x}=4$ To get rid of the radical, we square both sides.

x=4^2=16

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Example Question #1 : Solving Radical Equations

Solve for .

$\sqrt{x}=-5$

Possible Answers:

No answer

$\pm5$

$\pm25$

Correct answer:

No answer

Explanation:

To get rid of the radical, we need to square both sides. The issue is radicals don't generate negative numbers unless we talk about imaginary numbers. In this case, our answer choice should be no answer.

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Example Question #10 : Solving Radical Equations

Solve for .

$\sqrt{x+1}=5$

Possible Answers:

Correct answer:

Explanation:

$\sqrt{x+1}=5$ Square both sides to get rid of the radical.

x+1=5^2=25 Subtract on both sides.

x=24

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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 #1 : Solving Radical Equations

Example Question #2 : Solving Radical Equations

Example Question #3 : Solving Radical Equations

Example Question #4 : Solving Radical Equations

Example Question #5 : Solving Radical Equations

Example Question #6 : Solving Radical Equations

Example Question #7 : Solving Radical Equations

Example Question #1 : Solving And Graphing Radicals

Example Question #1 : Solving Radical Equations

Example Question #10 : Solving Radical Equations

All Algebra II Resources

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