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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3941 : Algebra Ii

Simplify: $10\sqrt{20}\cdot 30\sqrt{40}$

Possible Answers:

$6000\sqrt2$

$40+2\sqrt{10} +2\sqrt5$

$800\sqrt6$

$3000\sqrt2$

$20\sqrt{5}+60\sqrt{10}$

Correct answer:

$6000\sqrt2$

Explanation:

Multiply the integers and combine the radicals together by multiplication.

$10\sqrt{20}\cdot 30\sqrt{40} =300 \sqrt{800}$

Break up square root of 800 by common factors of perfect squares.

$300 \sqrt{800} = 300 \times \sqrt{100}\times \sqrt{4}\times \sqrt2$

Simplify the possible radicals.

$300\times 10 \times 2 \times \sqrt2 = 6000\sqrt2$

The answer is: $6000\sqrt2$

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Example Question #18 : Non Square Radicals

Evaluate: $5\sqrt{30}\cdot 4\sqrt{10}$

Possible Answers:

260

$200\sqrt3$

$20\sqrt{30}$

$200\sqrt6$

$6\sqrt5$

Correct answer:

$200\sqrt3$

Explanation:

Multiply the integers and the value of the square roots to combine as one radical.

$5\sqrt{30}\cdot 4\sqrt{10} = 20\sqrt{300}$

Simplify the radical. Use factors of perfect squares to simplify root 300.

$20\sqrt{300} = 20 \sqrt{100} \cdot \sqrt{3}= 20\cdot 10\cdot \sqrt3$

The answer is: $200\sqrt3$

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Example Question #41 : Understanding Radicals

Simplify: $12\sqrt{56}$

Possible Answers:

$48\sqrt{7}$

$29\sqrt5$

$24\sqrt{14}$

$\textup{This radical is already simplified.}$

$56\sqrt3$

Correct answer:

$24\sqrt{14}$

Explanation:

In order to simplify the radical, we will need to pull out common factors of possible perfect squares.

$12\sqrt{56} = 12\cdot \sqrt{8}\cdot \sqrt{7}=12\cdot \sqrt{4}\cdot \sqrt{2}\cdot \sqrt{7}$

The expression becomes:

$12\cdot 2\cdot \sqrt{2}\cdot \sqrt{7} = 24\sqrt{14}$

The radical 14 does not have any common factors of perfect squares.

The answer is: $24\sqrt{14}$

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Example Question #3944 : Algebra Ii

Simplify: $\sqrt{45}-\sqrt{80}$

Possible Answers:

$\sqrt{35}$

$-\sqrt{35}$

$-\sqrt5$

$-2\sqrt5$

Correct answer:

$-\sqrt5$

Explanation:

Simplify by factoring both radicals by perfect squares.

$\sqrt{45} = \sqrt{9}\cdot \sqrt{5} = 3\sqrt5$

$\sqrt{80} = \sqrt{16}\cdot \sqrt5 = 4\sqrt5$

Replace the terms.

$\sqrt{45}-\sqrt{80} = 3\sqrt5- 4\sqrt5$

Combine like-terms.

The answer is: $-\sqrt5$

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Example Question #3945 : Algebra Ii

Solve: $6\sqrt{3}\cdot 5\sqrt{5}\cdot2\sqrt{10}$

Possible Answers:

$300\sqrt6$

$6000\sqrt{15}$

$600\sqrt{15}$

600

$300\sqrt5$

Correct answer:

$300\sqrt6$

Explanation:

Multiply the integers outside of the radical.

$6\sqrt{3}\cdot 5\sqrt{5}\cdot2\sqrt{10} =60\sqrt{3}\cdot \sqrt{5}\cdot\sqrt{10}$

Multiply all the values inside the radicals to combine as one radical.

$60\sqrt{3}\cdot \sqrt{5}\cdot\sqrt{10} = 60\sqrt{150}$

Rewrite the radical using factors of perfect squares.

$60\sqrt{150} = 60\sqrt{25} \cdot \sqrt6 = 60(5)\sqrt6$

The answer is: $300\sqrt6$

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Example Question #3946 : Algebra Ii

Simplify: $10\sqrt{1000}\cdot \sqrt{\frac{1}{8}}$

Possible Answers:

$25\sqrt5$

$75\sqrt3$

$25\sqrt{2}$

$25\sqrt{10}$

$50\sqrt5$

Correct answer:

$50\sqrt5$

Explanation:

The values of the radicals can be combined my multiplication.

$10\sqrt{1000}\cdot \sqrt{\frac{1}{8}} = 10\sqrt{\frac{1000}{8}} = 10\sqrt{125}$

The value of $\sqrt{125}$ can be factored by using known perfect squares as factors.

$\sqrt{125} =\sqrt{25}\cdot \sqrt{5} = 5\sqrt5$

Replace the term.

$10\sqrt{125} = 10\cdot 5\sqrt5= 50\sqrt5$

The answer is: $50\sqrt5$

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Example Question #3947 : Algebra Ii

Simplify: $4\sqrt{8}\cdot \sqrt{27}\cdot\sqrt{50}$

Possible Answers:

$120\sqrt3$

$120\sqrt6$

$\textup{The answer is not given.}$

$240\sqrt3$

$240\sqrt6$

Correct answer:

$240\sqrt3$

Explanation:

Rewrite each radical as common factors using known perfect squares.

$4[\sqrt{4}\cdot \sqrt2]\cdot [\sqrt{9}\cdot \sqrt3]\cdot[\sqrt{25}\cdot \sqrt2]$

Simplify the expression.

$4[2\cdot \sqrt2]\cdot [3\cdot \sqrt3]\cdot[5\cdot \sqrt2]$

$8 \cdot 15\sqrt3 \cdot 2 = 240\sqrt3$

The answer is: $240\sqrt3$

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Example Question #1281 : Mathematical Relationships And Basic Graphs

True or false: $\sqrt[3]{52}$ is a radical expression in simplest form.

Possible Answers:

True

False

Correct answer:

True

Explanation:

A radical expression which is the th root of a constant is in simplest form if and only if, when the radicand is expressed as the product of prime factors, no factor appears or more times. Since $\sqrt[3]{52}$ is a cube, or third, root, find the prime factorization of 52, and determine whether any prime factor appears three or more times.

52 can be broken down as

$52 = 2 \times 26$

and further as

$52 = 2 \times 2 \times 13$

No prime factor appears three or more times, so $\sqrt[3]{52}$ is in simplest form.

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Example Question #3949 : Algebra Ii

Try without a calculator.

Simplify:

$\sqrt[3]{\frac{2}{7}}$

Possible Answers:

$\sqrt[3]{2}$

$\frac{\sqrt[3]{98}}{7}$

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

$\frac{\sqrt[3]{49}}{7}$

$\frac{2\sqrt[3]{49}}{7}$

Correct answer:

$\frac{\sqrt[3]{98}}{7}$

Explanation:

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

$\sqrt[3]{\frac{2}{7}}$

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

Since we are dealing with cube, or third, roots, rationalize the denominator by multiplying both halves of the fraction by the least cube-root radical expression that would eliminate the radical in the denominator. To do this, note that 7 is a prime number. Therefore, to get a perfect cube, it is necessary to multiply both halves by $\sqrt[3]{7 \cdot 7}$ , and apply the Product of Radicals Property. The reason for this is made more apparent below:

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

$=\frac{\sqrt[3]{2 \cdot 7 \cdot 7}}{\sqrt[3]{7 \cdot 7 \cdot 7}}$

$=\frac{\sqrt[3]{98}}{\sqrt[3]{7^{3}}}$

$=\frac{\sqrt[3]{98}}{7}$

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Example Question #3950 : Algebra Ii

Try without a calculator.

Simplify:

$\sqrt[3]{\frac{250}{8}}$

Possible Answers:

$\frac{ 5 \sqrt[3]{ 4 }}{2}$

$\frac{ \sqrt[3]{ 20}}{2}$

$\frac{ \sqrt[3]{ 25}}{2}$

$\frac{ 5 }{2}$

$\frac{ 5 \sqrt[3]{ 2 }}{2}$

Correct answer:

$\frac{ 5 \sqrt[3]{ 2 }}{2}$

Explanation:

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

$\sqrt[3]{\frac{250}{8}}$

$=\frac{\sqrt[3]{250}}{\sqrt[3]{8}}$

8 is a perfect cube - $8= 2^{3}$ - so the denominator can be simplified:

$\frac{\sqrt[3]{250}}{\sqrt[3]{2^{3}}}$

$=\frac{\sqrt[3]{250}}{2}$

To simplify the numerator, find the prime factorization of its radicand, 250, and look for any prime factors that appear three times:

$250 = 2 \times 125$

$= 2 \times 5 \times 25$

$= 2 \times 5 \times 5 \times 5$

$= 5 ^{3} \times 2$

5 appears three times, so the numerator can be simplified by way of the Product of Radicals Property:

$\frac{\sqrt[3]{5^{3} \times 2}}{2}$

$=\frac{\sqrt[3]{5^{3}} \times\sqrt[3]{ 2 }}{2}$

$=\frac{ 5 \sqrt[3]{ 2 }}{2}$

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #3941 : Algebra Ii

Example Question #18 : Non Square Radicals

Example Question #41 : Understanding Radicals

Example Question #3944 : Algebra Ii

Example Question #3945 : Algebra Ii

Example Question #3946 : Algebra Ii

Example Question #3947 : Algebra Ii

Example Question #1281 : Mathematical Relationships And Basic Graphs

Example Question #3949 : Algebra Ii

Example Question #3950 : Algebra Ii

All Algebra II Resources

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