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High School Math : Algebra II

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

Example Questions

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 #91 : Algebra Ii

Solve the equation for .

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

Possible Answers:

x=16

x=44.5

x=-44.5

x=40

x=-40

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 #42 : Radicals

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 #1 : Understanding Negative Exponents

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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Example Question #92 : Algebra Ii

Solve for :

$(x+5)^{-3} = -1$

Possible Answers:

Correct answer:

Explanation:

Raise both sides of the equation to the inverse power of to cancel the exponent on the left hand side of the equation.

$\rightarrow ((x+5)^{-3})^{-\frac{1}{3}} = (-1)^{-\frac{1}{3}}$

$\rightarrow x+5 = -1$

Subtract from both sides:

$\rightarrow (x+5) - 5 = (-1)-5$

$\rightarrow x = -6$

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Example Question #2 : Fractional Exponents

Convert the exponent to radical notation.

$x^{\frac{3}{7}}$

Possible Answers:

$\small \small \sqrt[3]{x^7}$

$\small \frac{1}{x^4}$

$\small \small \sqrt[7]{x^3}$

$\small \frac{x^3}{x^7}$

Correct answer:

$\small \small \sqrt[7]{x^3}$

Explanation:

Remember that exponents in the denominator refer to the root of the term, while exponents in the numerator can be treated normally.

$x^{\frac{a}{b}}=\sqrt[b]{x^a}$

$x^{\frac{3}{7}}=\sqrt[7]{x^3}$

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Example Question #1 : Understanding Exponents

Which of the following is equivalent to $64^{\frac{1}{2}}$ ?

Possible Answers:

$\frac{1}{32}$

Correct answer:

Explanation:

By definition, a number raised to the $\frac{1}{2}$ power is the same as the square root of that number.

Since the square root of 64 is 8, 8 is our solution.

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Example Question #1 : Fractional Exponents

Simplify the expression:

$\small (16^{\frac{1}{2}})(256^{\frac{3}{4}})$

Possible Answers:

1024

256

Correct answer:

256

Explanation:

Remember that fraction exponents are the same as radicals.

$\small 16^{\frac{1}{2}}=\sqrt{16}=4$

$256^{\frac{3}{4}}=\sqrt[4]{256^3}=64$

A shortcut would be to express the terms as exponents and look for opportunities to cancel.

$16^{\frac{1}{2}}=(4^2)^{\frac{1}{2}}=4$

$\small 256^{\frac{3}{4}}=(4^4)^{\frac{3}{4}}=4^3=64$

Either method, we then need to multiply to two terms.

$\small (4)(64)=256$

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Example Question #91 : Mathematical Relationships And Basic Graphs

Simplify the following expression.

$\frac{4x^3y^8z^2}{8x^6y^2z^4}$

Possible Answers:

$\frac{4y^6}{8x^3z^2}$

$\frac{y^4}{2x^2z^2}$

$\frac{4y^4}{8x^2z^2}$

$\frac{y^6}{2x^3z^2}$

Correct answer:

$\frac{y^6}{2x^3z^2}$

Explanation:

When dividing with exponents, the exponent in the denominator is subtracted from the exponent in the numerator. For example: $\frac{x^3}{x^6}= x^{3-6}=x^{-3}=\frac{1}{x^3}$ .

In our problem, each term can be treated in this manner. Remember that a negative exponent can be moved to the denominator.

$\frac{4x^3y^8z^2}{8x^6y^2z^4}=\frac{4}{8}x^{-3}y^6z^{-2}=\frac{4y^6}{8x^3z^2}$

Now, simplifly the numerals.

$\frac{4y^6}{8x^3z^2}=\frac{y^6}{2x^3z^2}$

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High School Math : Algebra II

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #6 : Solving And Graphing Radical Equations

Example Question #7 : Solving And Graphing Radical Equations

Example Question #91 : Algebra Ii

Example Question #42 : Radicals

Example Question #1 : Understanding Negative Exponents

Example Question #92 : Algebra Ii

Example Question #2 : Fractional Exponents

Example Question #1 : Understanding Exponents

Example Question #1 : Fractional Exponents

Example Question #91 : Mathematical Relationships And Basic Graphs

All High School Math Resources

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