How to find the common factor of square roots - ACT Math

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Question

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

Answer

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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Question

Solve for .

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

Answer

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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Question

Which of the following is equivalent to:

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

Answer

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

$\sqrt{210}+\sqrt{55}=\sqrt{2357}+\sqrt{511}$

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

$\sqrt{5}\sqrt{237}+\sqrt{5}\sqrt{11}$

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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Question

Simplify:

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

Answer

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

$\sqrt{15}-\sqrt{20}+\sqrt{35}=\sqrt{35}-\sqrt{45}+\sqrt{75}$

Now, this could be rewritten:

$\sqrt{35}-\sqrt{45}+\sqrt{75}=\sqrt{5}\sqrt{3}-\sqrt{5}\sqrt{4}+\sqrt{5}\sqrt{7}$

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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Question

Simplify:

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

Answer

Begin by factoring out the relevant squared data:

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

$x\sqrt{252}+x\sqrt{42}-\sqrt{92}+\sqrt{492}$

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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Question

Solve for :

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

Answer

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

$t\sqrt{47}-t\sqrt{97}=\sqrt{14}$

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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Question

Solve for :

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

Answer

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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Question

Solve for : $x\sqrt{45} + x\sqrt{108} = 12$

Answer

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

$x\sqrt{45} + x\sqrt{108} = 12$ ---> $x\sqrt{5 \cdot 9} + x\sqrt{3\cdot 36 } = 12$

Factor out the square root of each perfect square:

$x\sqrt{5 \cdot 9} + x\sqrt{3 \cdot 36} = 12$ ---> $3x\sqrt{5} + 6x\sqrt{3} = 12$

Next, factor out from each term on the left-hand side of the equation:

$3x\sqrt{5} + 3x\sqrt{12} = 12$ ---> $3x(\sqrt{5} +2\sqrt{3}) = 12$

Lastly, isolate :

$3x(\sqrt{5} +2\sqrt{3}) = 12$ ---> $x = \frac{4}{(\sqrt{5} +2\sqrt{3})}$

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Question

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

Answer

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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Question

Solve for :

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

Answer

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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Question

Solve for :

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

Answer

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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Question

Solve for :

$\frac{1}{2}x\sqrt[4]{9072} + x\sqrt[3]{54} = 18$

Answer

To begin solving this problem, find the greatest perfect square for all quantities under a radical. $3x\sqrt[4]{9072}$ might seem intimidating, but remember that raising even single-digit numbers to the fourth power creates huge numbers. In this case, is divisible by , a perfect fourth power.

$\frac{1}{2}x\sqrt[4]{9072} + x\sqrt[3]{54} = 18$ ---> $\frac{1}{2}x\sqrt[4]{7\cdot 1296} + x\sqrt[3]{2\cdot 27} = 18$

Pull the perfect terms out of each term on the left:

$\frac{1}{2}x\sqrt[4]{7\cdot 1296} + x\sqrt[3]{2\cdot 27} = 18$ ---> $3x\sqrt[4]{7} + 3x\sqrt[3]{2} = 18$

Next, factor out from the left-hand side:

$3x\sqrt[4]{7} + 3x\sqrt[3]{2} = 18$ ---> $3x(\sqrt[4]{7} + \sqrt[3]{2}) = 18$

Lastly, isolate , remembering to simplify the fraction where possible:

$3x(\sqrt[4]{7} + \sqrt[3]{2}) = 18$ ---> $x = \frac{6}{(\sqrt[4]{7} + \sqrt[3]{2})}$

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Question

Simplify: $\sqrt{384} + \sqrt{216}$

Answer

To start, begin pulling the largest perfect square you can out of each number:

$\sqrt{384} = \sqrt {4 \cdot 4 \cdot 4 \cdot 6 } = \sqrt {64 \cdot 6} = 8\sqrt 6$

$\sqrt {216} = \sqrt {6 \cdot 6 \cdot 6} = \sqrt {36 \cdot 6} = 6\sqrt 6$

So, $\sqrt{384} + \sqrt{216} = 8\sqrt 6 + 6\sqrt 6 = 14\sqrt{6}$ . You can just add the two terms together once they have a common radical.

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Question

Simplify: $\sqrt{6300} - \sqrt{700}$

Answer

Again here, it is easiest to recognize that both of our terms are divisible by , a prime number likely to appear in our final answer:

$\sqrt{6300} - \sqrt{700} = \sqrt{7\cdot 900} - \sqrt{7 \cdot 100}$

Now, simplify our perfect squares:

$\sqrt{7\cdot 900} - \sqrt{7 \cdot 100} = 30\sqrt 7 - 10 \sqrt 7$

Lastly, subtract our terms with a common radical:

$30\sqrt{7} - 10\sqrt {7} = 20\sqrt{7}$

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