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ACT Math : Basic Squaring / Square Roots

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Example Questions

Example Question #1 : How To Find The Common Factor Of Square Roots

$\sqrt{2016} + \sqrt{896} = ?$

Possible Answers:

75.0

$20\sqrt{14}$

$208\sqrt{14}$

$\sqrt{2919}$

54.0

Correct answer:

$20\sqrt{14}$

Explanation:

To solve the equation $\sqrt{2016} + \sqrt{896} = ?$ , we can first factor the numbers under the square roots.

$\sqrt{2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 7} + \sqrt{2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 7} = ?$

When a factor appears twice, we can take it out of the square root.

$2\cdot 2\cdot 3\sqrt{2\cdot 7} + 2\cdot 2\cdot 2\sqrt{2\cdot 7} = ?$

$12\sqrt{14} + 8 \sqrt{14} = ?$

Now the numbers can be added directly because the expressions under the square roots match.

$12\sqrt{14} + 8\sqrt{14} = 20\sqrt{14}$

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Example Question #1 : Factoring Common Factors Of Squares And Square Roots

Solve for .

$x\sqrt{96} + x\sqrt{150} = \sqrt{162}$

Possible Answers:

${\sqrt{3}}$

$\frac{1}{\sqrt{6}}$

$9\sqrt{2}$

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

$9\sqrt{12}$

Correct answer:

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

Explanation:

First, we can simplify the radicals by factoring.

$x\sqrt{96} + x\sqrt{150} = \sqrt{162}$

$x\sqrt{2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3} + x\sqrt{2\cdot 3\cdot 5\cdot 5} = \sqrt{2\cdot 3\cdot 3\cdot 3\cdot 3}$

$x\left (2\cdot 2 \right )\sqrt{2\cdot 3} + x\left ( 5\right )\sqrt{2\cdot 3} = \left (3\cdot 3 \right )\sqrt{2}$

$4x\sqrt{6} + 5x\sqrt{6} = 9\sqrt{2}$

Now, we can factor out the .

$x\left (4\sqrt{6} + 5\sqrt{6} \right ) = 9\sqrt{2}$

$x\left (9\sqrt{6} \right ) = 9\sqrt{2}$

Now divide and simplify.

$x = \frac{9\sqrt{2}}{9\sqrt{6} } = \frac{\sqrt{2}}{\sqrt{2\cdot 3}} = \frac{1}{\sqrt{3}}$

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Example Question #1 : Basic Squaring / Square Roots

Which of the following is equivalent to:

$\sqrt{210}+\sqrt{55}$ ?

Possible Answers:

$\sqrt{5}(\sqrt{42}+\sqrt{11})$

$5\sqrt{7}+\sqrt{11}$

$5\sqrt{462}$

$7\sqrt{30}+5\sqrt{11}$

$\sqrt{265}$

Correct answer:

$\sqrt{5}(\sqrt{42}+\sqrt{11})$

Explanation:

To begin with, factor out the contents of the radicals. This will make answering much easier:

$\sqrt{210}+\sqrt{55}=\sqrt{2*3*5*7}+\sqrt{5*11}$

They both have a common factor . This means that you could rewrite your equation like this:

$\sqrt{5}\sqrt{2*3*7}+\sqrt{5}\sqrt{11}$

This is the same as:

$\sqrt{5}\sqrt{42}+\sqrt{5}\sqrt{11}$

These have a common $\sqrt{5}$ . Therefore, factor that out:

$\sqrt{5}\sqrt{42}+\sqrt{5}\sqrt{11}=\sqrt{5}(\sqrt{42}+\sqrt{11})$

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Example Question #2 : Basic Squaring / Square Roots

Simplify:

$\sqrt{15}-\sqrt{20}+\sqrt{35}$

Possible Answers:

$\sqrt{2}(\sqrt{5}+2\sqrt{7})$

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

$\sqrt{5}(\sqrt{3}+\sqrt{7}-2)$

$\sqrt{5}(\sqrt{10}-2)$

$2\sqrt{15}+\sqrt{2}$

Correct answer:

$\sqrt{5}(\sqrt{3}+\sqrt{7}-2)$

Explanation:

These three roots all have a in common; therefore, you can rewrite them:

$\sqrt{15}-\sqrt{20}+\sqrt{35}=\sqrt{3*5}-\sqrt{4*5}+\sqrt{7*5}$

Now, this could be rewritten:

$\sqrt{3*5}-\sqrt{4*5}+\sqrt{7*5}=\sqrt{5}\sqrt{3}-\sqrt{5}\sqrt{4}+\sqrt{5}\sqrt{7}$

Now, note that $\sqrt{4}=2$

Therefore, you can simplify again:

$\sqrt{5}\sqrt{3}-\sqrt{5}\sqrt{4}+\sqrt{5}\sqrt{7}=\sqrt{5}\sqrt{3}-2\sqrt{5}+\sqrt{5}\sqrt{7}$

Now, that looks messy! Still, if you look carefully, you see that all of your factors have $\sqrt{5}$ ; therefore, factor that out:

$\sqrt{5}\sqrt{3}-2\sqrt{5}+\sqrt{5}\sqrt{7}=\sqrt{5}(\sqrt{3}-2+\sqrt{7})$

This is the same as:

$\sqrt{5}(\sqrt{3}+\sqrt{7}-2)$

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Example Question #1 : Factoring Common Factors Of Squares And Square Roots

Simplify:

$x\sqrt{50}+x\sqrt{8}-\sqrt{18}+\sqrt{98}$

Possible Answers:

22x

$11x\sqrt{2}$

$\sqrt{3}(5x-3)$

$\sqrt{2}(7x+4)$

$11x\sqrt{2}$

Correct answer:

$\sqrt{2}(7x+4)$

Explanation:

Begin by factoring out the relevant squared data:

$x\sqrt{50}+x\sqrt{8}-\sqrt{18}+\sqrt{98}$ is the same as

$x\sqrt{25*2}+x\sqrt{4*2}-\sqrt{9*2}+\sqrt{49*2}$

This can be simplified to:

$5x\sqrt{2}+2x\sqrt{2}-3\sqrt{2}+7\sqrt{2}$

Since your various factors contain square roots of , you can simplify:

$7x\sqrt{2}+4\sqrt{2}$

Technically, you can factor out a $\sqrt{2}$ :

$\sqrt{2}(7x+4)$

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Example Question #2 : How To Find The Common Factor Of Square Roots

Solve for :

$t\sqrt{28}-t\sqrt{63}=\sqrt{14}$

Possible Answers:

$\sqrt{14}$

$-\sqrt{14}$

$-\sqrt{2}$

Correct answer:

$-\sqrt{2}$

Explanation:

Begin by breaking apart the square roots on the left side of the equation:

$t\sqrt{4*7}-t\sqrt{9*7}=\sqrt{14}$

This can be rewritten:

$2t\sqrt{7}-3t\sqrt{7}=\sqrt{14}$

You can combine like terms on the left side:

$-t\sqrt{7}=\sqrt{14}$

Solve by dividing both sides by $-\sqrt{7}$ :

$t=\frac{-\sqrt{14}}{\sqrt{7}}$

This simplifies to:

$t=-\sqrt{2}$

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Example Question #6 : Basic Squaring / Square Roots

Solve for :

$x\sqrt{28} + x\sqrt{84} = 12$

Possible Answers:

$x = \frac{12}{(\sqrt{7} +\sqrt{21})}$

$x = \frac{6}{(\sqrt{14} +\sqrt{21})}$

$x = \frac{6}{(\sqrt{7} -\sqrt{21})}$

$x = \frac{6}{(\sqrt{7} +\sqrt{21})}$

$x = \frac{\sqrt{7}}{(\sqrt{7} +\sqrt{21})}$

Correct answer:

$x = \frac{6}{(\sqrt{7} +\sqrt{21})}$

Explanation:

To begin solving this problem, find the greatest common perfect square for all quantities under a radical.

$x\sqrt{28} + x\sqrt{84} = 12$ ---> $x\sqrt{4 \cdot 7} + x\sqrt{4 \cdot 21} = 12$

Pull out of each term on the left:

$x\sqrt{4 \cdot 7} + x\sqrt{4 \cdot 21} = 12$ ---> $2x\sqrt{7} + 2x\sqrt{21} = 12$

Next, factor out from the left-hand side:

$2x\sqrt{7} + 2x\sqrt{21} = 12$ ---> $2x(\sqrt{7} +\sqrt{21}) = 12$

Lastly, isolate :

$2x(\sqrt{7} +\sqrt{21}) = 12$ ---> $x = \frac{6}{(\sqrt{7} +\sqrt{21})}$

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Example Question #7 : Basic Squaring / Square Roots

Solve for : $x\sqrt{125} + 5x\sqrt{70} = \sqrt{5}$

Possible Answers:

$x = \frac{5}{1+10\sqrt{3}}$

$x = \frac{1}{5+10\sqrt{3}}$

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

$x = \frac{5}{1+\sqrt{12}}$

$x = \frac{5}{5\sqrt{12}}$

Correct answer:

$x = \frac{1}{5+10\sqrt{3}}$

Explanation:

Solving this one is tricky. At first glance, we have no common perfect square to work with. But since each term can produce the quantity $\sqrt{5}$ , let's start there:

$x\sqrt{125} + 5x\sqrt{70} = \sqrt{5}$ ---> $x\sqrt{25 \cdot 5} + 5x\sqrt{12 \cdot 5} = \sqrt{5}$

Simplify the first term:

$x\sqrt{25 \cdot 5} + 5x\sqrt{12 \cdot 5} = \sqrt{5}$ ---> $5x\sqrt{5} + 5x\sqrt{12 \cdot 5} = \sqrt{5}$

Divide all terms by $\sqrt{5}$ to simplify,

$5x\sqrt{5} + 5x\sqrt{12 \cdot 5} = \sqrt{5}$ ---> $5x + 5x\sqrt{12} = 1$

Next, factor out from the left-hand side:

$5x + 5x\sqrt{12} = 1$ ---> $5x(1+ \sqrt{12}) = 1$

Isolate by dividing by $5(1+\sqrt{12})$ and simplifying:

$5x(1+ \sqrt{12}) = 1$ ---> $x = \frac{1}{5+5\sqrt{12}}$

Last, simplify the denominator:

$x = \frac{1}{5+5\sqrt{12}}$ ----> $x = \frac{1}{5+10\sqrt{3}}$

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Example Question #34 : Arithmetic

Solve for :

$x\sqrt{23} - x\sqrt{69} = 12\sqrt{115}$

Possible Answers:

$x = \frac{12\sqrt{23}}{1-\sqrt{3}}$

$x = \frac{1-\sqrt{3}}{12\sqrt{5}}$

$x = \frac{1-\sqrt{5}}{12\sqrt{23}}$

$x = \frac{12\sqrt{3}}{1-\sqrt{5}}$

$x = \frac{12\sqrt{5}}{1-\sqrt{3}}$

Correct answer:

$x = \frac{12\sqrt{5}}{1-\sqrt{3}}$

Explanation:

Right away, we notice that $\sqrt{23}$ is a prime radical, so no simplification is possible. Note, however, that both other radicals are divisible by $\sqrt{23}$ .

Our first step then becomes simplifying the equation by dividing everything by $\sqrt{23}$ :

$x\sqrt{23} - x\sqrt{69} = 12\sqrt{115}$ ---> $x-x\sqrt{3} = 12\sqrt{5}$

Next, factor out from the left-hand side:

$x-x\sqrt{3} = 12\sqrt{5}$ ---> $x(1-\sqrt{3}) = 12\sqrt{5}$

Lastly, isolate :

$x(1-\sqrt{3}) = 12\sqrt{5}$ ---> $x = \frac{12\sqrt{5}}{1-\sqrt{3}}$

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Example Question #8 : How To Find The Common Factor Of Square Roots

Solve for :

$-4x\sqrt{\frac{1}{4}} + \frac{2}{3}x\sqrt{27} = \sqrt{15}$

Possible Answers:

$x= \frac{\sqrt{15}}{2+2\sqrt{3}}$

$x= \frac{\sqrt{15}}{-2+2\sqrt{3}}$

$x= \frac{\sqrt{2}}{-3+2\sqrt{15}}$

$x= \frac{\sqrt{5}}{-2+2\sqrt{3}}$

$x= \frac{\sqrt{15}}{-2-3\sqrt{2}}$

Correct answer:

$x= \frac{\sqrt{15}}{-2+2\sqrt{3}}$

Explanation:

Once again, there are no common perfect squares under the radical, but with some simplification, the equation can still be solved for :

$-4x\sqrt{\frac{1}{4}} + \frac{2}{3}x\sqrt{27} = \sqrt{15}$ ---> $(\frac{1}{2}\cdot -4x) + (3 \cdot\frac{2}{3}x\sqrt{3}) = \sqrt{15}$

Simplify:

$(\frac{1}{2}\cdot -4x) + (3 \cdot\frac{2}{3}x\sqrt{3}) = \sqrt{15}$ ---> $(-2x) + (2x\sqrt{3}) = \sqrt{15}$

Factor out from the left-hand side:

$(-2x) + (2x\sqrt{3}) = \sqrt{15}$ ---> $2x(-1 + \sqrt{3}) = \sqrt{15}$

Lastly, isolate :

$2x(-1 + \sqrt{3}) = \sqrt{15}$ ---> $x= \frac{\sqrt{15}}{-2+2\sqrt{3}}$

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ACT Math : Basic Squaring / Square Roots

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #1 : How To Find The Common Factor Of Square Roots

Example Question #1 : Factoring Common Factors Of Squares And Square Roots

Example Question #1 : Basic Squaring / Square Roots

Example Question #2 : Basic Squaring / Square Roots

Example Question #1 : Factoring Common Factors Of Squares And Square Roots

Example Question #2 : How To Find The Common Factor Of Square Roots

Example Question #6 : Basic Squaring / Square Roots

Example Question #7 : Basic Squaring / Square Roots

Example Question #34 : Arithmetic

Example Question #8 : How To Find The Common Factor Of Square Roots

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