How to find the common factors of squares

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PSAT Math › How to find the common factors of squares

Questions 1 - 3

1

Simplify the radical:

$\sqrt{320}$

$8\sqrt{5}$

$4\sqrt{20}$

$4\sqrt{5}$

$8\sqrt{20}$

$\sqrt{320}$

Explanation

$\sqrt{320}=\sqrt{2\times 160}=\sqrt{2\times 2\times 80}=2\sqrt{80}=2\sqrt{2\times 40}=2\sqrt{2\times 2\times 20}=2\times 2\times \sqrt{20}=4\sqrt{2\times 10}=4\sqrt{2\times 2\times 5}=4\times 2\times \sqrt{5}=8\sqrt{5}$

2

Solve for $\dpi{100} x$ :

$x\sqrt{45}+x\sqrt{72}=\sqrt{18}$

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

$x=\frac{\sqrt{2}}{\sqrt{5}}+\frac{1}{2}$

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

$x=\sqrt{9}$

$x=3$

Explanation

$x\sqrt{45}+x\sqrt{72}=\sqrt{18}$

Notice how all of the quantities in square roots are divisible by 9

$x\sqrt{9\times 5}+x\sqrt{9\times 8}=\sqrt{9\times 2}$

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

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

$3x\sqrt{5}+6x\sqrt{2}=3\sqrt{2}$

$x(3\sqrt{5}+6\sqrt{2})=3\sqrt{2}$

$x=\frac{3\sqrt{2}}{3\sqrt{5}+6\sqrt{2}}$

Simplifying, this becomes

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

3

If m and n are postive integers and 4m = 2n, what is the value of m/n?

1/2

2

4

8

16

Explanation

22 = 4. Also, following the rules of exponents, 41 = 1.
One can therefore say that m = 1 and n = 2.
The question asks to solve for m/n. Since m = 1 and n = 2, m/n = 1/2.

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