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**Bacterius** There are some very important formulas you need to learn by heart in order to understand square roots. They are :

$\displaystyle \sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$

$\displaystyle \sqrt{a \times b^2} = b \times \sqrt{a}$

Note that the second one can be proved from the first one :

$\displaystyle \sqrt{a \times b^2} = \sqrt{a} \times \sqrt{b^2} = \sqrt{a} \times b = b \times \sqrt{a}$

This way, if the expression under the square root can be broken into factors, and that some factors can be written as squares, then they can be "extracted" from the square root.

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Let's take, $\displaystyle \sqrt{48}$ for the example. You want to simplify it.

First, break $\displaystyle 48$ into its prime factors, to make it easier to spot squares that you can remove from the square root.

Note that $\displaystyle 48 = 24 \times 2 = 2 \times 12 \times 2$ $\displaystyle = 2 \times 2 \times 6 \times 2$ $\displaystyle = 2 \times 2 \times 2 \times 3 \times 2$ $\displaystyle = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3$

So, $\displaystyle \sqrt{48} = \sqrt{2^4 \times 3}$.

Now, remember $\displaystyle 2^4 = (2^2)^2$, so :

$\displaystyle \sqrt{48} = \sqrt{(2^2)^2 \times 3}$

And you see that there is a square. You can remove it :

$\displaystyle \sqrt{48} = \sqrt{(2^2)^2} \sqrt{3}$

And the $\displaystyle (2^2)^2$ square root is easily simplified :

$\displaystyle \sqrt{48} = \sqrt{16} \sqrt{3}$

$\displaystyle \sqrt{48} = 4 \sqrt{3}$

Do you understand better now ?