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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #1551 : Mathematical Relationships And Basic Graphs

Simplify: $\frac{7\sqrt7}{\sqrt{\frac{1}{7}}}$

Possible Answers:

$\sqrt{14}$

$\frac{1}{7}$

$\frac{1}{49}$

Correct answer:

Explanation:

The radical $\sqrt{\frac{1}{7}}$ can be rewritten as:

$\frac{\sqrt1}{\sqrt7} = \frac{1}{\sqrt7}$

Rationalize the denominator by multiplying both the top and bottom by the denominator.

$\frac{1}{\sqrt7} \cdot \frac{\sqrt7}{\sqrt7} =\frac{\sqrt7}{7}$

Rewrite the given question.

$\frac{7\sqrt7}{\sqrt{\frac{1}{7}}} = \frac{7\sqrt7}{\frac{\sqrt7}{7}}$

Simplify this complex fraction by rewriting it as a division sign.

$\frac{7\sqrt7}{\frac{\sqrt7}{7}} = 7\sqrt7 \div \frac{\sqrt7}{7}$

Change the division sign to multiplication and take the reciprocal of the second term.

$7\sqrt7 \div \frac{\sqrt7}{7} = 7\sqrt7 \times \frac{7}{\sqrt7} =49$

The answer is:

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Example Question #4211 : Algebra Ii

Rationalize: $\frac{9}{\sqrt{20}}$

Possible Answers:

$\frac{3\sqrt{2}}{4}$

$\frac{9\sqrt{2}}{4}$

$\frac{3\sqrt{5}}{10}$

$\frac{9\sqrt{5}}{10}$

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

Correct answer:

$\frac{9\sqrt{5}}{10}$

Explanation:

We can factor the square root twenty before rationalizing the value so that the final term would not be as difficult to factor.

Factor $\sqrt{20}$ using factors of perfect squares.

$\sqrt{20}= \sqrt{4}\times \sqrt5 = 2\sqrt5$

The expression becomes:

$\frac{9}{\sqrt{20}} = \frac{9}{2\sqrt{5}}$

Instead of multiplying $\sqrt{20}$ on both the top and bottom, we will multiply $\sqrt5$ on the top and bottom.

$\frac{9}{2\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{9\sqrt{5}}{10}$

The answer is: $\frac{9\sqrt{5}}{10}$

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Example Question #222 : Simplifying Radicals

Try without a calculator.

Simplify:

$\sqrt{\frac{5}{8}}$

Possible Answers:

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

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

$\frac{\sqrt{10}}{4}$

None of the other choices gives the correct response.

$\frac{\sqrt{5}}{4}$

Correct answer:

$\frac{\sqrt{10}}{4}$

Explanation:

First, apply the Quotient of Radicals Property to split the radical into a numerator and a denominator:

$\sqrt{\frac{5}{8}}$

$= \frac{\sqrt{5}}{\sqrt{8}}$

Simplify the denominator by applying the Product of Radicals Property, as follows:

$= \frac{\sqrt{5}}{\sqrt{4} \cdot \sqrt{2}}$

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

Rationalize the denominator by multiplying both halves of the fraction by $\sqrt{2}$ :

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

Apply the Product of Radicals property in the numerator:

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

$= \frac{\sqrt{10}}{4}$ ,

the correct response.

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Example Question #4212 : Algebra Ii

Try without a calculator.

Simplify:

$\sqrt{\frac{12}{7}}$

Possible Answers:

$\frac{3\sqrt{14} }{7}$

None of these

$\frac{ \sqrt{42} }{7}$

$\frac{2 \sqrt{21} }{7}$

$\frac{6\sqrt{7} }{7}$

Correct answer:

$\frac{2 \sqrt{21} }{7}$

Explanation:

First, apply the Quotient of Radicals Property to split the radical into a numerator and a denominator:

$\sqrt{\frac{12}{7}}$

$= \frac{\sqrt{12}}{\sqrt{7}}$

Simplify the numerator by applying the Product of Radicals Property, as follows:

$\frac{\sqrt{4 \cdot 3 }}{\sqrt{7}}$

$= \frac{\sqrt{4} \cdot \sqrt{3}}{\sqrt{7}}$

$= \frac{2 \sqrt{3}}{\sqrt{7}}$

Rationalize the denominator by multiplying both halves of the fraction by $\sqrt{7}$ :

$\frac{2 \sqrt{3} \cdot \sqrt{7} }{\sqrt{7} \cdot \sqrt{7}}$

Apply the Product of Radicals Property in the numerator:

$\frac{2 \sqrt{3 \cdot 7} }{7}$

$= \frac{2 \sqrt{21} }{7}$

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Example Question #1 : Solving And Graphing Radicals

Solve for :

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

Possible Answers:

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

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

None of the other responses is correct.

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

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

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 #1 : Solving Radical Equations

Solve the following radical equation.

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

Possible Answers:

$13\sqrt{2}$

$8\sqrt{2}$

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

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

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=2

x=-10

x=10

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=10

x=50

$x=2\sqrt{5}$

$x=\sqrt{10}$

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 #4 : Solving Radical Equations

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

Possible Answers:

507.3158

1,899.5

9,999.999

6,561

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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Algebra II : Mathematical Relationships and Basic Graphs

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1551 : Mathematical Relationships And Basic Graphs

Example Question #4211 : Algebra Ii

Example Question #222 : Simplifying Radicals

Example Question #4212 : Algebra Ii

Example Question #1 : Solving And Graphing Radicals

Example Question #1 : Solving Radical Equations

Example Question #3 : Solving Radical Equations

Example Question #4 : Solving Radical Equations

Example Question #5 : Solving Radical Equations

Example Question #4 : Solving Radical Equations

All Algebra II Resources

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