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High School Math : High School Math

Study concepts, example questions & explanations for High School Math

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8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Solving Radical Equations And Inequalities

Solve for :

$\sqrt{x^2+12x+36} =4$

Possible Answers:

$x = \{-2, 0\}$

x = -2

$x = \{-10, -6, -2\}$

$x = \left \{ -10, -2\right \}$

x = -10

Correct answer:

$x = \left \{ -10, -2\right \}$

Explanation:

To solve for in the equation $\sqrt{x^2+12x+36} =4$

Square both sides of the equation

$\rightarrow (\sqrt{x^2+12x+36})^2 =(4)^2$

$\rightarrow x^2+12x+36 =16$

Set the equation equal to by subtracting the constant from both sides of the equation.

$\rightarrow (x^2+12x+36) - 16 =(16) -16$

$\rightarrow x^2+12x+20 =0$

Factor to find the zeros:

$\rightarrow (x+10)(x+2) = 0$

This gives the solutions

$x = \left \{ -10, -2\right \}$ .

Verify that these work in the original equation by substituting them in for . This is especially important to do in equations involving radicals to ensure no imaginary numbers (square roots of negative numbers) are created.

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Example Question #1 : Solving And Graphing Radical Equations

Solve the following radical expression:

$\sqrt{3x+1}+2=7$

Possible Answers:

Correct answer:

Explanation:

Begin by subtracting from each side of the equation:

$\sqrt{3x+1}+2=7$

$\sqrt{3x+1}=5$

Now, square the equation:

$(\sqrt{3x+1}=5)^2$

3x+1=25

Solve the linear equation:

3x=24

x=8

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Example Question #3 : Solving And Graphing Radical Equations

Solve the following radical expression:

$\sqrt{2x+1}-2=\sqrt{2x-7}$

Possible Answers:

Correct answer:

Explanation:

Begin by squaring both sides of the equation:

$\sqrt{2x+1}-2=\sqrt{2x-7}$

$(\sqrt{2x+1}-2)^2=(\sqrt{2x-7})^2$

$(2x+1)-4\sqrt{2x+1}+4=2x-7$

Combine like terms:

$-4\sqrt{2x+1}=-12$

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

Once again, square both sides of the equation:

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

2x+1=9

Solve the linear equation:

2x=8

x=4

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Example Question #3 : Solving Radical Equations And Inequalities

Solve the following radical expression:

$\sqrt{4x+29}=(x+6)$

Possible Answers:

No real solutions

Correct answer:

Explanation:

Begin by squaring both sides of the equation:

$\sqrt{4x+29}=(x+6)$

$(\sqrt{4x+29})^2=(x+6)^2$

4x+29=x^2+12x+36

Now, combine like terms:

x^2+8x+7=0

Factor the equation:

(x+7)(x+1)=0

x=-1,-7

However, when plugging in the values, does not work. Therefore, there is only one solution:

x=-1

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Example Question #5 : Solving And Graphing Radical Equations

Solve the following radical expression:

$\sqrt{y+12}+1=\sqrt{y+21}$

Possible Answers:

Correct answer:

Explanation:

Begin by squaring both sides of the equation:

$\sqrt{y+12}+1=\sqrt{y+21}$

$(\sqrt{y+12}+1)^2=(\sqrt{y+21})^2$

$(y+12)+2\sqrt{y+12}+1=y+21$

Now, combine like terms and simplify:

$2\sqrt{y+12}=8$

$\sqrt{y+12}=4$

Once again, take the square of both sides of the equation:

$(\sqrt{y+12}=4)^2$

y+12=16

Solve the linear equation:

y=4

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Example Question #6 : Solving And Graphing Radical Equations

Solve the following radical expression:

$\sqrt{5+2x}=x-5$

Possible Answers:

Correct answer:

Explanation:

Begin by taking the square of both sides:

$\sqrt{5+2x}=x-5$

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

5+2x=x^2-10x+25

Combine like terms:

x^2-12x+20=0

Factor the equation and solve:

(x-10)(x-2)=0

x=10,2

However, when plugging in the values, does not work. Therefore, there is only one solution:

x=10

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Example Question #7 : Solving And Graphing Radical Equations

Solve the following radical expression:

$\sqrt{x+11}-x=-9$

Possible Answers:

Correct answer:

Explanation:

To solve the radical expression, begin by subtracting from each side of the equation:

$\sqrt{x+11}-x=-9$

$\sqrt{x+11}=x-9$

Now, square both sides of the equation:

$(\sqrt{x+11})^2=(x-9)^2$

x+11=x^2-18x+81

Combine like terms:

x^2-19x+70=0

Factor the expression and solve:

(x-14)(x-5)=0

x=14,5

However, when plugged into the original equation, does not work because the radical cannot be negative. Therefore, there is only one solution:

x=14

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Example Question #1281 : High School Math

Solve the equation for .

$\sqrt{4x+9}-8=5$

Possible Answers:

x=16

x=-40

x=44.5

x=40

x=-44.5

Correct answer:

x=40

Explanation:

$\small \sqrt{4x+9}-8=5$

Add to both sides.

$\small \sqrt{4x+9}=13$

Square both sides.

$\small (\sqrt{4x+9})^2=13^2$

$\small 4x+9=169$

Isolate .

$\small 4x=160$

$\small x=40$

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Example Question #2 : Solving Radical Equations And Inequalities

Solve for :

$\sqrt[3]{y+1}=3$

Possible Answers:

Correct answer:

Explanation:

Begin by cubing both sides:

$\sqrt[3]{y+1}=3$

$(\sqrt[3]{y+1}=3)^3$

y+1=27

Now we can easily solve:

y=26

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Example Question #91 : Algebra Ii

Which of the following is equivalent to $3^{-2}$ ?

Possible Answers:

$\frac{1}{6}$

$\frac{1}{9}$

Correct answer:

$\frac{1}{9}$

Explanation:

By definition,

$b^{-x} = \frac{1}{b^{x}}$ .

In our problem, b = 3 and x = 2 .

Then, we have $\frac{1}{3^{2}} = \frac{1}{9}$ .

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High School Math : High School Math

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Solving Radical Equations And Inequalities

Example Question #1 : Solving And Graphing Radical Equations

Example Question #3 : Solving And Graphing Radical Equations

Example Question #3 : Solving Radical Equations And Inequalities

Example Question #5 : Solving And Graphing Radical Equations

Example Question #6 : Solving And Graphing Radical Equations

Example Question #7 : Solving And Graphing Radical Equations

Example Question #1281 : High School Math

Example Question #2 : Solving Radical Equations And Inequalities

Example Question #91 : Algebra Ii

All High School Math Resources

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