GRE Math : How to simplify square roots

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Example Question #1 : How To Simplify Square Roots

Simplify the following: (√(6) + √(3)) / √(3)

Possible Answers:

3√(2)

√(3)

√(2) + 1

None of the other answers

Correct answer:

√(2) + 1

Explanation:

Begin by multiplying top and bottom by √(3):

(√(18) + √(9)) / 3

Note the following:

√(9) = 3

√(18) = √(9 * 2) = √(9) * √(2) = 3 * √(2)

Therefore, the numerator is: 3 * √(2) + 3. Factor out the common 3: 3 * (√(2) + 1)

Rewrite the whole fraction:

(3 * (√(2) + 1)) / 3

Simplfy by dividing cancelling the 3 common to numerator and denominator: √(2) + 1

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Example Question #2 : How To Simplify Square Roots

what is

√0.0000490

Possible Answers:

0.00007

0.007

0.07

Correct answer:

0.007

Explanation:

easiest way to simplify: turn into scientific notation

√0.0000490= √4.9 X 10^-5

finding the square root of an even exponent is easy, and 49 is a perfect square, so we can write out an improper scientific notation:

√4.9 X 10^-5 = √49 X 10^-6

√49 = 7; √10^-6 = 10^-3 this is equivalent to raising 10^-6 to the 1/2 power, in which case all that needs to be done is multiply the two exponents: 7 X 10^-3= 0.007

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Example Question #3 : How To Simplify Square Roots

Simplify: $\sqrt{576}$ $\displaystyle \sqrt{576}$

Possible Answers:

$10\sqrt{12}$ $\displaystyle 10\sqrt{12}$

$\displaystyle 24$

$12\sqrt{3}$ $\displaystyle 12\sqrt{3}$

$12\sqrt{6}$ $\displaystyle 12\sqrt{6}$

$\displaystyle 34$

Correct answer:

$\displaystyle 24$

Explanation:

In order to take the square root, divide 576 by 2.

$\dpi{100} \sqrt{576}= \sqrt{2}\sqrt{288}=\sqrt{2}\sqrt{2}\sqrt{144}=\sqrt{4}\sqrt{144}=2\cdot 12=24$ $\displaystyle \dpi{100} \sqrt{576}= \sqrt{2}\sqrt{288}=\sqrt{2}\sqrt{2}\sqrt{144}=\sqrt{4}\sqrt{144}=2\cdot 12=24$

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Example Question #4 : How To Simplify Square Roots

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

Possible Answers:

$\frac{2}{3}$ $\displaystyle \frac{2}{3}$

$\frac{4}{9}$ $\displaystyle \frac{4}{9}$

$\frac{2}{81}$ $\displaystyle \frac{2}{81}$

$\frac{8}{81}$ $\displaystyle \frac{8}{81}$

$\frac{4}{81}$ $\displaystyle \frac{4}{81}$

Correct answer:

$\frac{2}{3}$ $\displaystyle \frac{2}{3}$

Explanation:

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

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

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

$\displaystyle =$ $\frac{2}{3}$ $\displaystyle \frac{2}{3}$

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Example Question #22 : Arithmetic

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

Possible Answers:

$4\sqrt{5x}$ $\displaystyle 4\sqrt{5x}$

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

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

$2x\sqrt{10}$ $\displaystyle 2x\sqrt{10}$

Correct answer:

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

Explanation:

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

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

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Example Question #5 : How To Simplify Square Roots

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

Possible Answers:

$3\sqrt{5}$ $\displaystyle 3\sqrt{5}$

$\displaystyle 9$

$7.5\sqrt{10}$ $\displaystyle 7.5\sqrt{10}$

$5\sqrt{3}$ $\displaystyle 5\sqrt{3}$

Correct answer:

$5\sqrt{3}$ $\displaystyle 5\sqrt{3}$

Explanation:

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

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Example Question #12 : How To Simplify Square Roots

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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Example Question #6 : How To Simplify Square Roots

What is $\sqrt{50}$ $\displaystyle \sqrt{50}$?

Possible Answers:

$5\sqrt{2}$ $\displaystyle 5\sqrt{2}$

$10\sqrt{2}$ $\displaystyle 10\sqrt{2}$

$\displaystyle 5$

$\displaystyle 10$

$2\sqrt{5}$ $\displaystyle 2\sqrt{5}$

Correct answer:

$5\sqrt{2}$ $\displaystyle 5\sqrt{2}$

Explanation:

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

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Example Question #13 : How To Simplify Square Roots

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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Example Question #14 : How To Simplify Square Roots

Which of the following is the most simplified form of:

$\sqrt{468}$ $\displaystyle \sqrt{468}$

Possible Answers:

$2\sqrt{117}$ $\displaystyle 2\sqrt{117}$

$6\sqrt{13}$ $\displaystyle 6\sqrt{13}$

$17\sqrt{2}$ $\displaystyle 17\sqrt{2}$

$\sqrt{468}$ $\displaystyle \sqrt{468}$

$4\sqrt{29}$ $\displaystyle 4\sqrt{29}$

Correct answer:

$6\sqrt{13}$ $\displaystyle 6\sqrt{13}$

Explanation:

First find all of the prime factors of 468 $\displaystyle 468$

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

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

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GRE Math : How to simplify square roots

Study concepts, example questions & explanations for GRE Math

All GRE Math Resources

Example Questions

Example Question #1 : How To Simplify Square Roots

Example Question #2 : How To Simplify Square Roots

Example Question #3 : How To Simplify Square Roots

Example Question #4 : How To Simplify Square Roots

Example Question #22 : Arithmetic

Example Question #5 : How To Simplify Square Roots

Example Question #12 : How To Simplify Square Roots

Example Question #6 : How To Simplify Square Roots

Example Question #13 : How To Simplify Square Roots

Example Question #14 : How To Simplify Square Roots

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