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Algebra II : Non-Square Radicals

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Example Question #21 : Non Square Radicals

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

Possible Answers:

$-\sqrt{35}$

$\sqrt{35}$

$-2\sqrt5$

$-\sqrt5$

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 #22 : Non Square Radicals

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

Possible Answers:

$300\sqrt6$

$300\sqrt5$

600

$600\sqrt{15}$

$6000\sqrt{15}$

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 #23 : Non Square Radicals

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

Possible Answers:

$75\sqrt3$

$25\sqrt{2}$

$25\sqrt{10}$

$25\sqrt5$

$50\sqrt5$

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 #24 : Non Square Radicals

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

Possible Answers:

$120\sqrt6$

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

$120\sqrt3$

$240\sqrt6$

$240\sqrt3$

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 #41 : Understanding Radicals

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 #41 : Understanding Radicals

Try without a calculator.

Simplify:

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

Possible Answers:

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

$\sqrt[3]{2}$

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

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

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

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 #27 : Non Square Radicals

Try without a calculator.

Simplify:

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

Possible Answers:

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

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

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

$\frac{ 5 }{2}$

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

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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Example Question #21 : Non Square Radicals

Rationalize the denominator and simplify:

$\frac{1}{6+\sqrt{27}}$

Possible Answers:

$\frac{6+\sqrt{27}}{6-\sqrt{27}}$

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

$\frac{6-3\sqrt{3}}{9}$

$\frac{\sqrt{27}-6}{9}$

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

Correct answer:

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

Explanation:

To rationalize a denominator, multiply all terms by the conjugate. In this case, the denominator is $6+\sqrt{27}$ , so its conjugate will be $6-\sqrt{27}$ .

So we multiply: $\frac{1}{6+\sqrt{27}} \times \frac{6-\sqrt{27}}{6-\sqrt{27}} = \frac{6-\sqrt{27}}{36-27}$ .

After simplifying, we get $\frac{6-\sqrt{27}}{36-27} = \frac{3\cdot2-3\sqrt{3}}{9}=\frac{2-\sqrt{3}}{3}$ .

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Algebra II : Non-Square Radicals

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #21 : Non Square Radicals

Example Question #22 : Non Square Radicals

Example Question #23 : Non Square Radicals

Example Question #24 : Non Square Radicals

Example Question #41 : Understanding Radicals

Example Question #41 : Understanding Radicals

Example Question #27 : Non Square Radicals

Example Question #21 : Non Square Radicals

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