GRE Math : How to divide square roots

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Example Question #11 : How To Divide Square Roots

Solve for :

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

Possible Answers:

$\frac{1}{16}$

Correct answer:

Explanation:

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.

x=16

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Example Question #12 : How To Divide Square Roots

Simplify:

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

Possible Answers:

$\sqrt{\cos \theta }$

$\sqrt{\sin \theta }$

$\tan\theta*\sqrt{\sin \theta }$

$\tan^2\theta*\sqrt{\sin \theta }$

$\tan^2\theta*\sqrt{\cos\theta }$

Correct answer:

$\tan^2\theta*\sqrt{\sin \theta }$

Explanation:

$\tan\theta$ 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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Example Question #13 : How To Divide Square Roots

Simplify.

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

Possible Answers:

$\sqrt{5}$

$5+2\sqrt{5}$

$\frac{5-2\sqrt{5}}{3}$

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

$\frac{\sqrt{5}}{3}$

Correct answer:

$5+2\sqrt{5}$

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.

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

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GRE Math : How to divide square roots

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Example Question #11 : How To Divide Square Roots

Example Question #12 : How To Divide Square Roots

Example Question #13 : How To Divide Square Roots

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