GRE Math : How to divide square roots

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Example Question #41 : Arithmetic

Which of the following is equal to $\sqrt{81}/\sqrt{6}$

Possible Answers:

$3\sqrt{6}/2$

$3\sqrt{3}$

$2\sqrt{3}$

$\sqrt{3}/2$

$3\sqrt{6}$

Correct answer:

$3\sqrt{6}/2$

Explanation:

$\sqrt{81}/\sqrt{6}=9/\sqrt{6}$ We then multiply our fraction by $\sqrt{6}/\sqrt{6}$ because we cannot leave a radical in the denominator. This gives us $9\sqrt{6}/6$ . Finally, we can simplify our fraction, dividing out a 3, leaving us with $3\sqrt{6}/2$

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Example Question #42 : Arithmetic

Simplify:

$\frac{\sqrt{343x^5}}{\sqrt{49x^3}}$

Possible Answers:

$\frac{x}{7}$

$7\sqrt{x}$

$\frac{7}{x}$

$x\sqrt{7}$

Correct answer:

$x\sqrt{7}$

Explanation:

Let's combine the two radicals into one radical and simplify.

$\frac{\sqrt{343x^5}}{\sqrt{49x^3}}=\sqrt{\frac{343x^5}{49x^3}}=\sqrt{7x^2}$

Remember, when dividing exponents of same base, just subtract the power.

The final answer is $x\sqrt{7}$ .

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Example Question #43 : Arithmetic

Solve for :

$\frac{1}{\sqrt{x}}=4$

Possible Answers:

$\frac{1}{4}$

$\pm \frac{1}{16}$

$\frac{1}{16}$

Correct answer:

$\frac{1}{16}$

Explanation:

If we multiplied top and bottom by $\sqrt{x}$ , we would get nowhere, as this would result: $\frac{1}{\sqrt{x}}*\frac{\sqrt{x}}{\sqrt{x}}=\frac{\sqrt{x}}{x}$ . Instead, let's cross-multiply.

$\frac{1}{\sqrt{x}}=\frac{4}{1}$

$4\cdot \sqrt{x}=1\cdot 1$

$4\sqrt{x}=1$

Then, square both sides to get rid of the radical.

$(4\sqrt{x})^2=1^2$

16x=1

Divide both sides by .

$x=\frac{1}{16}$

The reason the negative is not an answer is because a negative value in a radical is an imaginary number.

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Example Question #1 : How To Divide Square Roots

Rationalize the denominator:

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

Possible Answers:

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

$\frac{2}{5}$

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

$\frac{5}{2}$

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

Correct answer:

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

Explanation:

We don't want to have radicals in the denominator. To get rid of radicals, just multiply top and bottom by that radical.

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

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Example Question #1 : How To Divide Square Roots

Simplify:

$\frac{\sqrt{250}}{\sqrt{10}}$

Possible Answers:

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

$\sqrt{10}$

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

Correct answer:

Explanation:

There are two methods we can use to simplify this fraction:

Method 1:

Factor the numerator:

$\frac{\sqrt{250}}{\sqrt{10}}=\frac{\sqrt{25}\cdot\sqrt{10}}{\sqrt{10}}=\sqrt{25}=5$

Remember, we need to factor out perfect squares.

Method 2:

You can combine the fraction into one big square root.

$\frac{\sqrt{250}}{\sqrt{10}}=\sqrt{\frac{250}{10}}$

Then, you can simplify the fraction.

$\sqrt{\frac{250}{10}}=\sqrt{25}=5$

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Example Question #46 : Arithmetic

Simplify:

$\frac{\sqrt{21}}{\sqrt{20}}$

Possible Answers:

$\frac{21}{20}$

$\frac{\sqrt{21}}{10}$

$\frac{\sqrt{105}}{10}$

$\frac{\sqrt{105}}{50}$

$\frac{105}{10}$

Correct answer:

$\frac{\sqrt{105}}{10}$

Explanation:

Let's factor the square roots.

$\frac{\sqrt{21}}{\sqrt{20}}=\frac{\sqrt{3}\cdot\sqrt{7}}{\sqrt{4}\cdot \sqrt{5}}=\frac{\sqrt{3}\cdot \sqrt{7}}{2\cdot \sqrt{5}}$

Then, multiply the numerator and the denominator by $\sqrt{5}$ to get rid of the radical in the denominator.

$\frac{\sqrt{3}\cdot \sqrt{7}\cdot \sqrt{5}}{2\cdot \sqrt{5}\cdot \sqrt{5}}=\frac{\sqrt{105}}{10}$

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Example Question #47 : Arithmetic

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

Possible Answers:

$\frac{1}{3}$

$\sqrt{3}$

$\frac{1}{\sqrt{3}}$

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

Correct answer:

$\frac{1}{\sqrt{3}}$

Explanation:

We can definitely eliminate some answer choices. and $\frac{1}{3}$ don't make sense because we have an irrational number. Next, let's multiply the numerator and denominator of $\frac{\sqrt{3}}{3}$ by $\sqrt{3}$ . When we simplify radical fractions, we try to eliminate radicals, but here, we are going to go backwards.

$\frac{\sqrt{3}\cdot \sqrt{3}}{3\cdot \sqrt{3}}=\frac{3}{3\sqrt{3}}=\frac{1}{\sqrt{3}}$

$\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$ , so $\frac{1}{\sqrt{3}}$ is the answer.

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Example Question #1 : How To Divide Square Roots

Rationalize the denominator and simplify:

$\frac{\sqrt{8}+\sqrt{12}}{\sqrt{6}}$

Possible Answers:

$\frac{\sqrt{48}+\sqrt{72}}{6}$

$\frac{\sqrt{15}}{3}$

$\frac{5}{3}$

$\frac{\sqrt{30}}{3}$

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

Correct answer:

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

Explanation:

We don't want to have radicals in the denominator. To get rid of radicals, just multiply the numerator and the denominator by that radical.

$\frac{\sqrt{8}+\sqrt{12}}{\sqrt{6}}\cdot \frac{\sqrt{6}}{\sqrt{6}}=$ $\frac{\sqrt{48}+\sqrt{72}}{6}$

Remember to distribute the radical in the numerator when multiplying.

This may be the answer; however, the numerator can be simplified. Let's factor out the squares.

$\frac{\sqrt{48}+\sqrt{72}}{6}=\frac{\sqrt{16}\cdot \sqrt{3}+\sqrt{36}\cdot \sqrt{2}}{6}=\frac{4\sqrt{3}+6\sqrt{2}}{6}$

Finally, if we factor out a , we get:

$\frac{2}{2}\cdot \frac{2\sqrt{3}+3\sqrt{2}}{3}=\frac{2\sqrt{3}+3\sqrt{2}}{3}$

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Example Question #49 : Arithmetic

Simplify:

$\sqrt{\frac{5}{6}}-\sqrt{\frac{7}{8}}$

Possible Answers:

$\sqrt{\frac{35}{48}}\cdot -\frac{1}{24}$

$\frac{\sqrt{105}}{12}$

$\frac{4\sqrt{30}}{24}-\frac{3\sqrt{56}}{24}$

$\frac{2\sqrt{14}-3\sqrt{30}}{12}$

$\frac{2\sqrt{30}-3\sqrt{14}}{12}$

Correct answer:

$\frac{2\sqrt{30}-3\sqrt{14}}{12}$

Explanation:

Let's get rid of the radicals in the denominator of each individual fraction.

$\sqrt{\frac{5}{6}}-\sqrt{\frac{7}{8}}=\sqrt{\frac{5\cdot 6}{6\cdot 6}}-\sqrt{\frac{7\cdot 8}{8\cdot 8}}=\frac{\sqrt{30}}{6}-\frac{\sqrt{56}}{8}$

Then find the least common denominator of the fractions, which is , and multiply them so that they each have a denominator of .

$\frac{4\sqrt{30}}{24}-\frac{3\sqrt{56}}{24}$

We can definitely simplify the numerator in the right fraction by factoring out a perfect square of .

$\frac{4\sqrt{30}}{24}-\frac{3\sqrt{56}}{24}=\frac{4\sqrt{30}}{24}-\frac{3\sqrt{4}*\sqrt{14}}{24}=\frac{4\sqrt{30}}{24}-\frac{6\sqrt{14}}{24}$

Finally, we can factor out a :

$\frac{2}{2}\cdot \frac{2\sqrt{30}-3\sqrt{14}}{12}=\frac{2\sqrt{30}-3\sqrt{14}}{12}$

That's the final answer.

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Example Question #50 : Arithmetic

Simplify:

$\frac{1+\sqrt{2}}{1-\sqrt{2}}$

Possible Answers:

$-3-2\sqrt{2}$

$-5\sqrt{2}$

$3+2\sqrt{2}$

$-3+2\sqrt{2}$

$-\sqrt{2}$

Correct answer:

$-3-2\sqrt{2}$

Explanation:

To get rid of the radical, we need to multiply by the conjugate. The conjugate uses the opposite sign and multiplying by it will let us rationalize the denominator in this problem. The goal is getting an expression of a^2-b^2 in which we are taking the differences of two squares.

$\frac{1+\sqrt{2}}{1-\sqrt{2}}\cdot \frac{{1+\sqrt{2}}}{{1+\sqrt{2}}}=\frac{1+\sqrt{2}+\sqrt{2}+2}{1-2}=\frac{3+2\sqrt{2}}{-1}$

This answer is the same as $-3-2\sqrt{2}$ . Remember to distribute the negative sign.

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GRE Math : How to divide square roots

Study concepts, example questions & explanations for GRE Math

All GRE Math Resources

Example Questions

Example Question #41 : Arithmetic

Example Question #42 : Arithmetic

Example Question #43 : Arithmetic

Example Question #1 : How To Divide Square Roots

Example Question #1 : How To Divide Square Roots

Example Question #46 : Arithmetic

Example Question #47 : Arithmetic

Example Question #1 : How To Divide Square Roots

Example Question #49 : Arithmetic

Example Question #50 : Arithmetic

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