How to divide square roots - GRE Quantitative Reasoning

Card 0 of 13

Question

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

Answer

$\sqrt{81}/\sqrt{6}=9/\sqrt{6}$ 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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Question

Rationalize the denominator:

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

Answer

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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Question

Simplify:

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

Answer

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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Question

Simplify:

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

Answer

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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Question

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

Answer

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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Question

Rationalize the denominator and simplify:

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

Answer

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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Question

Simplify:

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

Answer

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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Question

Simplify:

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

Answer

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 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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Question

Simplify.

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

Answer

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 in which we are taking the differences of two squares.

$\frac{\sqrt{5}}{\sqrt{5}-2}\cdot \frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{5+2\sqrt{5}}{5-4}=\frac{5+2\sqrt{5}}{1}=5+2\sqrt{5}$

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Question

Solve for :

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

Answer

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.

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

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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Question

Simplify:

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

Answer

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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Question

Solve for :

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

Answer

To get rid of the radical, multiply top and bottom by $\sqrt{x}$ .

$\frac{x}{\sqrt{x}}=\frac{x}{\sqrt{x}}\cdot \frac{\sqrt{x}}{\sqrt{x}}=\frac{x\sqrt{x}}{x}=\sqrt{x}$

$\sqrt{x}=4$

Square both sides.

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Question

Simplify:

$\frac{\sin \theta }{\cot \theta }*\frac{\sqrt{\tan \theta }}{\sqrt{\cos \theta} }$

Answer

$\tan\theta$ is "opposite over adjacent," or $\frac{\sin \theta }{\cos \theta }$ . (This is because $tan\theta=\frac{sin\theta}{cos\theta}=\frac{\frac{opposite}{hypotenuse}}{\frac{adjacent}{hypotenuse}}=\frac{opposite}{adjacent}$ .)

Let's substitute $\frac{\sin \theta }{\cos \theta }$ into the equation for $\tan\theta$ and multiply the numerator and the denominator by $\sqrt{cos \theta }$ .

$\sqrt{\frac{\tan \theta }{\cos \theta }}=\sqrt{\frac{\frac{\sin \theta }{\cos \theta }}{\cos \theta }}\cdot \frac{\sqrt{\cos \theta }}{\sqrt{\cos \theta }}=\frac{\sqrt{\sin \theta }}{\cos \theta }$

Now we can multiply the result by $\frac{\sin\theta}{\cot\theta}$ :

$\frac{\sqrt{\sin \theta }}{\cos \theta }\cdot \frac{\sin \theta }{\cot \theta }$

$\cot \theta$ is the inverse of $\tan\theta$ , or $\frac{\cos \theta }{\sin \theta }$ :

$\frac{\sqrt{\sin \theta }}{\cos \theta }\cdot \frac{\sin \theta }{\frac{\cos\theta}{\sin\theta}}=\frac{\sin\theta}{\cos\theta}\cdot \frac{\sqrt{\sin\theta}}{\cot\theta}$

We can reduce the resulting $\frac{\sin \theta }{\cos \theta }$ to $\tan\theta$ :

$\frac{\sin\theta}{\cos\theta}\cdot \frac{\sqrt{\sin\theta}}{\cot\theta}=\frac{\tan \theta \cdot \sqrt{\sin \theta }}{\cot \theta }$

By multiplying top and bottom by $\cot \theta$ , we can cancel out the $\tan\theta$ in the numerator:

$\frac{\tan \theta \cdot \sqrt{\sin \theta }}{\cot \theta }\cdot \frac{\cot\theta}{\cot\theta}=\frac{\sqrt{\sin\theta}}{\cot^2\theta}$

The resulting fraction can be simplified:

$\frac{\sqrt{\sin\theta}}{\cot^2\theta}=tan^2\theta\cdot\sqrt{sin\theta}$

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