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

Solve for .

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

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

$9\sqrt{2}$

${\sqrt{3}}$

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

$9\sqrt{12}$

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.

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

Solve for :

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

$-\sqrt{2}$

$-\sqrt{14}$

$\sqrt{14}$

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:

$t=-\sqrt{2}$

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

$20\sqrt{14}$

$208\sqrt{14}$

$\sqrt{2919}$

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.

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

Solve for :

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

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

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

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

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

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

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 :

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

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

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

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

$x = \frac{4}{(\sqrt{5} +\sqrt{12})}$

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

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

Explanation

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 :

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

Simplify:

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

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

$11x\sqrt{2}$

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

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 $\sqrt{2}$ :

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

Solve for :

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

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

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

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

$x = \frac{3}{(\sqrt[8]{7} + \sqrt[6]{2})}$

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

Explanation

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:

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

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

$14\sqrt{6}$

$8\sqrt{8}$

$16\sqrt{7}$

$8\sqrt{14}$

$12\sqrt{7}$

Explanation

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.

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

$20 \sqrt{7}$

$15\sqrt{21}$

$7\sqrt{13}$

$13\sqrt{13}$

$8\sqrt{15}$

Explanation

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:

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

Solve for :

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

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

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

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

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

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

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 :

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

How to find the common factor of square roots

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