GRE Math : Basic Squaring / Square Roots

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Example Questions

Example Question #4 : Simplifying Square Roots

Simplify $(\frac{16}{81})^{1/4}$ .

Possible Answers:

$\frac{4}{81}$

$\frac{4}{9}$

$\frac{2}{3}$

$\frac{2}{81}$

$\frac{8}{81}$

Correct answer:

$\frac{2}{3}$

Explanation:

$(\frac{16}{81})^{1/4}$

$\frac{16^{1/4}}{81^{1/4}}$

$\frac{(2\cdot 2\cdot 2\cdot 2)^{1/4}}{(3\cdot 3\cdot 3\cdot 3)^{1/4}}$

$\frac{2}{3}$

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Example Question #5 : How To Simplify Square Roots

Simplfy the following radical $\sqrt{20x^{2}}$ .

Possible Answers:

$4\sqrt{5x}$

$2\sqrt{5x^{2}}$

$2x\sqrt{10}$

$2x\sqrt{5}$

Correct answer:

$2x\sqrt{5}$

Explanation:

You can rewrite the equation as $\sqrt{20x^2}=(x)\sqrt{5} \cdot \sqrt{4}$ .

This simplifies to $2x\sqrt{5}$ .

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Example Question #21 : Arithmetic

Which of the following is equal to $\sqrt{75}$ ?

Possible Answers:

$3\sqrt{5}$

$5\sqrt{3}$

$7.5\sqrt{10}$

Correct answer:

$5\sqrt{3}$

Explanation:

√75 can be broken down to √25 * √3. Which simplifies to 5√3.

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Example Question #1 : How To Simplify Square Roots

Simplify $\sqrt{a^{3}b^{4}c^{5}}$ .

Possible Answers:

$ab^{2}c^{2}\sqrt{ac}$

$a^{2}bc\sqrt{bc}$

$a^{2}b^{2}c\sqrt{ab}$

$a^{2}b^{2}c^{2}\sqrt{bc}$

$a^{2}bc^{2}\sqrt{ac}$

Correct answer:

$ab^{2}c^{2}\sqrt{ac}$

Explanation:

Rewrite what is under the radical in terms of perfect squares:

$x^{2}=x\cdot x$

$x^{4}=x^{2}\cdot x^{2}$

$x^{6}=x^{3}\cdot x^{3}$

Therefore, $\sqrt{a^{3}b^{4}c^{5}}= \sqrt{a^{2}a^{1}b^{4}c^{4}c^{1}}=ab^{2}c^{2}\sqrt{ac}$ .

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Example Question #1 : Simplifying Square Roots

What is $\sqrt{50}$ ?

Possible Answers:

$2\sqrt{5}$

$10\sqrt{2}$

$5\sqrt{2}$

Correct answer:

$5\sqrt{2}$

Explanation:

We know that 25 is a factor of 50. The square root of 25 is 5. That leaves $\sqrt{2}$ which can not be simplified further.

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Example Question #1 : Exponents And Roots

Which of the following is equivalent to $\frac{x + \sqrt{3}}{3x + \sqrt{2}}$ ?

Possible Answers:

$\frac{3x^{2} + 3x\sqrt{2} + x\sqrt{3} +\sqrt{6}}{9x^{2} - 2}$

$\frac{4x + \sqrt{5}}{3x + 2}$

$\frac{3x^{2} + \sqrt{6}}{3x - 2}$

$\frac{3x^{2} -x \sqrt{2} + 3x\sqrt{3} - \sqrt{6}}{9x^{2} - 2}$

$\frac{3x^{2} - \sqrt{6}}{9x^{2} + 2}$

Correct answer:

$\frac{3x^{2} -x \sqrt{2} + 3x\sqrt{3} - \sqrt{6}}{9x^{2} - 2}$

Explanation:

Multiply by the conjugate and the use the formula for the difference of two squares:

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

$\frac{x + \sqrt{3}}{3x + \sqrt{2}}\cdot \frac{3x - \sqrt{2}}{3x - \sqrt{2}}$

$\frac{3x^{2} -x \sqrt{2} + 3x\sqrt{3} - \sqrt{6}}{(3x)^{2} - (\sqrt{2})^{2}}$

$\frac{3x^{2} -x \sqrt{2} + 3x\sqrt{3} - \sqrt{6}}{9x^{2} - 2}$

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Example Question #2 : How To Simplify Square Roots

Which of the following is the most simplified form of:

$\sqrt{468}$

Possible Answers:

$6\sqrt{13}$

$\sqrt{468}$

$2\sqrt{117}$

$17\sqrt{2}$

$4\sqrt{29}$

Correct answer:

$6\sqrt{13}$

Explanation:

First find all of the prime factors of 468

$468=6\ast78=6\ast6\ast13=2\ast3\ast2\ast3\ast13$

So $\sqrt{468}=\sqrt{2\ast2\ast3\ast3\ast13}=2\ast3\sqrt{13}=6\sqrt{13}$

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Example Question #2 : Exponents And Roots

What is $\sqrt{432}$ equal to?

Possible Answers:

$12\sqrt{3}$

$6\sqrt{3}$

$12\sqrt{12}$

$6\sqrt{4}$

$144\sqrt{3}$

Correct answer:

$12\sqrt{3}$

Explanation:

$\sqrt{432}$

1. We know that 432=(144)(3) , which we can separate under the square root:

$\sqrt{144\cdot 3}$

2. 144 can be taken out since it is a perfect square: $12\cdot12=144$ . This leaves us with:

$12\sqrt{3}$

This cannot be simplified any further.

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Example Question #1 : How To Find The Common Factor Of Square Roots

Which of the following is equivalent to:

$\sqrt{210}+\sqrt{55}$ ?

Possible Answers:

$7\sqrt{30}+5\sqrt{11}$

$\sqrt{265}$

$5\sqrt{462}$

$5\sqrt{7}+\sqrt{11}$

$\sqrt{5}(\sqrt{42}+\sqrt{11})$

Correct answer:

$\sqrt{5}(\sqrt{42}+\sqrt{11})$

Explanation:

To begin with, factor out the contents of the radicals. This will make answering much easier:

$\sqrt{210}+\sqrt{55}=\sqrt{2*3*5*7}+\sqrt{5*11}$

They both have a common factor . This means that you could rewrite your equation like this:

$\sqrt{5}\sqrt{2*3*7}+\sqrt{5}\sqrt{11}$

This is the same as:

$\sqrt{5}\sqrt{42}+\sqrt{5}\sqrt{11}$

These have a common $\sqrt{5}$ . Therefore, factor that out:

$\sqrt{5}\sqrt{42}+\sqrt{5}\sqrt{11}=\sqrt{5}(\sqrt{42}+\sqrt{11})$

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Example Question #1 : Factoring Common Factors Of Squares And Square Roots

Simplify:

$\sqrt{15}-\sqrt{20}+\sqrt{35}$

Possible Answers:

$\sqrt{5}(\sqrt{10}-2)$

$2\sqrt{15}+\sqrt{2}$

$\sqrt{7}-3\sqrt{5}$

$\sqrt{2}(\sqrt{5}+2\sqrt{7})$

$\sqrt{5}(\sqrt{3}+\sqrt{7}-2)$

Correct answer:

$\sqrt{5}(\sqrt{3}+\sqrt{7}-2)$

Explanation:

These three roots all have a in common; therefore, you can rewrite them:

$\sqrt{15}-\sqrt{20}+\sqrt{35}=\sqrt{3*5}-\sqrt{4*5}+\sqrt{7*5}$

Now, this could be rewritten:

$\sqrt{3*5}-\sqrt{4*5}+\sqrt{7*5}=\sqrt{5}\sqrt{3}-\sqrt{5}\sqrt{4}+\sqrt{5}\sqrt{7}$

Now, note that $\sqrt{4}=2$

Therefore, you can simplify again:

$\sqrt{5}\sqrt{3}-\sqrt{5}\sqrt{4}+\sqrt{5}\sqrt{7}=\sqrt{5}\sqrt{3}-2\sqrt{5}+\sqrt{5}\sqrt{7}$

Now, that looks messy! Still, if you look carefully, you see that all of your factors have $\sqrt{5}$ ; therefore, factor that out:

$\sqrt{5}\sqrt{3}-2\sqrt{5}+\sqrt{5}\sqrt{7}=\sqrt{5}(\sqrt{3}-2+\sqrt{7})$

This is the same as:

$\sqrt{5}(\sqrt{3}+\sqrt{7}-2)$

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GRE Math : Basic Squaring / Square Roots

Study concepts, example questions & explanations for GRE Math

All GRE Math Resources

Example Questions

Example Question #4 : Simplifying Square Roots

Example Question #5 : How To Simplify Square Roots

Example Question #21 : Arithmetic

Example Question #1 : How To Simplify Square Roots

Example Question #1 : Simplifying Square Roots

Example Question #1 : Exponents And Roots

Example Question #2 : How To Simplify Square Roots

Example Question #2 : Exponents And Roots

Example Question #1 : How To Find The Common Factor Of Square Roots

Example Question #1 : Factoring Common Factors Of Squares And Square Roots

All GRE Math Resources

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