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ACT Math : Arithmetic

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Example Question #32 : Basic Squaring / Square Roots

Simplify the following:

$\small \sqrt{132} - \sqrt{297}$

Possible Answers:

$-\sqrt{33}$

$-10\sqrt{21}$

$-2\sqrt{99}$

$\small -\sqrt{165}$

$\textup{The answer is not a real number}$

Correct answer:

$-\sqrt{33}$

Explanation:

Begin simplifying by breaking apart the square roots in question. Thus, you know:

$\small \small \small \sqrt{132} - \sqrt{297} = \sqrt{4*33} - \sqrt{9 * 33} = 2\sqrt{33} - 3\sqrt{33}$

Now, with square roots, you can combine factors just as if a given root were a variable. So, just as $\small 2x-3x=-x$ , so too does $\small 2\sqrt{33} - 3\sqrt{33}=-\sqrt{33}$ .

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Example Question #61 : Arithmetic

What is $5\sqrt{25}$ divided by $\sqrt{25}$ ?

Possible Answers:

$5\sqrt{5}$

$25\sqrt{5}$

$\sqrt{5}$

Correct answer:

Explanation:

Upon dividing by $\sqrt{25}$ , the $\sqrt{25}$ in the numerator cancels out, and the only number remaining is :

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

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Example Question #1131 : Act Math

Simplify $\sqrt{112} \div \sqrt{63}$

Possible Answers:

$\frac{4}{3}$

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

$\frac{16\sqrt{7}}{9}$

$\frac{4\sqrt{7}}{3}$

$\frac{16}{9}$

Correct answer:

$\frac{4}{3}$

Explanation:

Find the greatest factor that is a perfect square for 112 and 63. For 112, the factors are 16 and 7, thus $\sqrt{112} = \sqrt{16 \times 7}$ . For 63, the factors are 9 and 7, thus $\sqrt{63} = \sqrt{9 \times 7}$ . Simplifying these terms will give $\frac{4\sqrt{7}}{3\sqrt{7}}$ . Cancel out the $\sqrt{7}$ results in $\frac{4}{3}$ .

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Example Question #3 : How To Divide Square Roots

Simplify the following expression:

$\frac{\sqrt[3]{343}}{\sqrt{343}}$

Possible Answers:

343

Cannot be simplified any further

$\frac{\sqrt{7}}{7}$

$\sqrt{7}$

Correct answer:

$\frac{\sqrt{7}}{7}$

Explanation:

Although it may seem as though we cannot do anything to this expression due to our numerator and denominator having different indices, there is in fact some real simplifying to be done here.

To begin, our numerator can be evaluated, because 343 is in fact a perfect cube:

$\sqrt[3]{343}=7$

This fact helps us out with our denominator as well. Our original equation becomes the following:

$\frac{7}{\sqrt{7*7*7}}$

Then, we can pull out two of the sevens on the bottom and cancel them like so:

$\frac{7}{\sqrt{7*7}*\sqrt{7}}=\frac{7}{7\sqrt7}=\frac{1}{\sqrt{7}}$

We may seem to be done, but if you look, you will not see our solution among the answer choices. That is because we need to rationalize the denominator. We should never have a square root in our denominator. To remedy this, we do a fairly simple move that won't change the value of our fraction, but just the form it is in:

$\frac{1}{\sqrt{7}}*\frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{\sqrt{7}*\sqrt{7}}=\frac{\sqrt{7}}{7}}}$

Now we'll have no issue finding our answer choice!

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Example Question #4 : How To Divide Square Roots

Simplify:

$\small \frac{\sqrt{84}}{\sqrt{2079}}$

Possible Answers:

$\small \frac{2\sqrt{11}}{33}$

$\small \small \frac{2\sqrt{14}}{11}$

$\small \small \frac{3\sqrt{14}}{11}$

$\small \small \frac{9\sqrt{77}}{14}$

$\small \small \frac{3\sqrt{11}}{7}$

Correct answer:

$\small \frac{2\sqrt{11}}{33}$

Explanation:

Division of square roots is easy, since you can combine the roots and treat it like any other fraction. Thus, you can convert our fraction as follows:

$\small \small \frac{\sqrt{84}}{\sqrt{2079}} = \sqrt{\frac{84}{2079}}$

Next, you begin to reduce the fraction:

$\small \sqrt{\frac{84}{2079}}= \sqrt{\frac{2*2*3*7}{3*3 *3*7*11}}$

This reduces to:

$\small \small \sqrt{\frac{2*2}{3*3 *11}}$

Now, break this apart again into:

$\small \small \frac{\sqrt{4}}{\sqrt{9*11}}$ , which is $\small \small \small \frac{2}{3\sqrt{11}}$

Finish by rationalizing the denominator:

$\small \frac{2}{3\sqrt{11}} = \frac{2}{3\sqrt{11}} * \frac{\sqrt{11}}{\sqrt{11}}=\frac{2\sqrt{11}}{33}$

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Example Question #61 : Arithmetic

Simplify:

$\small \frac{\sqrt{1155}}{\sqrt{560}}$

Possible Answers:

$\small \frac{2\sqrt{77}}{3}$

$\small \frac{\sqrt{33}}{4}$

$\small \frac{3\sqrt{55}}{2}$

$\small 2\sqrt{55}$

$\small \frac{7\sqrt{10}}{6}$

Correct answer:

$\small \frac{\sqrt{33}}{4}$

Explanation:

Division of square roots is easy, since you can combine the roots and treat it like any other fraction. Thus, you can convert our fraction as follows:

$\small \frac{\sqrt{1155}}{\sqrt{560}} = \sqrt{\frac{1155}{560}}$

Next, you begin to reduce the fraction:

$\small \small \sqrt{\frac{1155}{560}} = \sqrt{\frac{3*5*7*11}{2*2*2*2*5*7}}$

This reduces to:

$\small \sqrt{\frac{3*11}{2*2*2*2}}$

Now, break this apart again into:

$\small \frac{\sqrt{33}}{\sqrt{16}}$ , which is $\small \frac{\sqrt{33}}{4}$

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Example Question #1 : How To Simplify Square Roots

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

Possible Answers:

$3\sqrt{5}$

$7.5\sqrt{10}$

$5\sqrt{3}$

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:

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

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

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

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

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

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 #11 : How To Simplify Square Roots

What is $\sqrt{50}$ ?

Possible Answers:

$5\sqrt{2}$

$10\sqrt{2}$

$2\sqrt{5}$

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 : How To Simplify Square Roots

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

Possible Answers:

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

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

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

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

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

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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ACT Math : Arithmetic

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #32 : Basic Squaring / Square Roots

Example Question #61 : Arithmetic

Example Question #1131 : Act Math

Example Question #3 : How To Divide Square Roots

Example Question #4 : How To Divide Square Roots

Example Question #61 : Arithmetic

Example Question #1 : How To Simplify Square Roots

Example Question #1 : How To Simplify Square Roots

Example Question #11 : How To Simplify Square Roots

Example Question #1 : How To Simplify Square Roots

All ACT Math Resources

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