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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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Example Questions

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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Example Question #111 : Solving And Graphing Radicals

If $\sqrt{2x+3}-5 = -2$ , then ?

Possible Answers:

-3

-4

Correct answer:

Explanation:

We will begin by adding 5 to both sides of the equation.

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

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

Next, let's square both sides to eliminate the radical.

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

2x+3 = 9

Finally, we can solve this like a simple two-step equation. Subtract 3 from both sides of the equation.

2x+3-3 = 9-3

2x=6

Now, divide each side by 2.

$\frac{2x}{2}=\frac{6}{2}$

x=3

Finally, check the solution to make sure that it results in a true statement.

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

$\sqrt{9}=3$

3=3

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Example Question #1666 : Mathematical Relationships And Basic Graphs

Solve for :

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

Possible Answers:

16, 9

Correct answer:

Explanation:

When working with radicals, a helpful step is to square both sides of an equation so that you can remove the radical sign and deal with a more classic linear or quadratic equation. But of course if you were to simply square both sides first here, you would still end up with radical signs, as were you to FOIL the left side you wouldn't eliminate the radicals. So a good first step is to add $2\sqrt{x}$ and subtract from both sides so that you get:

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

Now when you square both sides, you'll eliminate the radical on the right-hand side and yield:

x^2-6x+9=4x

Then when you subtract from both sides to set up a quadratic equalling zero, you have a factorable quadratic:

x^2-10x+9=0

This factors to:

(x-9)(x-1)=0

Which would seem to yield solutions of x=9 and x=1 . However, when you're solving for quadratics it's always important to plug your solutions back into the original equation to check for extraneous solutions. Here if you plug in x=9 the math holds: $9-2{\sqrt{9}}=3$ because $2\sqrt{9}=2(3)=6$ , and 9-6=3 .

But if you plug in x=1 , you'll see that the original equation is not satisfied: $1-3\neq 2\sqrt{1}$ because does not equal . Therefore the only proper answer is x=9 .

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Example Question #1667 : Mathematical Relationships And Basic Graphs

Solve for :

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

Possible Answers:

$\frac{-1 \pm \sqrt{89}}{2 }$

No solution

Correct answer:

Explanation:

To solve, first square both sides:

$\sqrt{x-3}^2 = (x-5)^2$ squaring the left side just givs x - 3. To square the left side, use the distributite property and multiply (x-5)(x-5) :

x - 3 = x^2 - 10x + 25 This is a quadratic, we just need to combine like terms and get it equal to 0

0 = x^2 -11x + 28 now we can solve using the quadratic formula:

$x = \frac{11 \pm \sqrt{11^2 - 4*1*28}}{2 } = \frac{11 \pm \sqrt {121 - 112}}{2 } = \frac{11 \pm \sqrt{9 }}{2} = \frac{11 \pm 3 }{2}$

This gives us 2 potential answers:

$x = \frac{11 + 3 }{2 } = \frac{14 }{2} = 7$ and

$x = \frac{11 - 3 }{ 2 } = \frac{8}{2} = 4$

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Example Question #1 : Graph Square Root, Cube Root, And Piecewise Functions

State the domain of the function:

$y=\sqrt{4x+3}$

Possible Answers:

$x\geq\frac{3}{4}$

$x\leq0$

$x\geq -\frac{4}{3}$

$x\geq0$

$x\geq-\frac{3}{4}$

Correct answer:

$x\geq-\frac{3}{4}$

Explanation:

Since the expression under the radical cannot be negative,

$4x+3\geq0$ .

Solve for x:

$x\geq-\frac{3}{4}$

This is the domain, or possible values, for the function.

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

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

Example Question #111 : Solving And Graphing Radicals

Example Question #1666 : Mathematical Relationships And Basic Graphs

Example Question #1667 : Mathematical Relationships And Basic Graphs

Example Question #1 : Graph Square Root, Cube Root, And Piecewise Functions

All Algebra II Resources

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