# Simplifying difference of square roots

I trying to review for calculus and I can't figure out how to do $\sqrt{200} - \sqrt{32}$

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$\sqrt {100 \cdot 2}-\sqrt {16 \cdot 2}$ – pedja Dec 17 '11 at 17:38
is $\sqrt{2\times 100} - \sqrt{2\times 16}$ of any help? – user20266 Dec 17 '11 at 17:38
I think I get it that leaves me with $10*\sqrt{2} - 4*\sqrt{2}$ and through some math property they are allowed to cancel out leaving me $6*\sqrt{2}$ – user138246 Dec 17 '11 at 17:40
You mean the distributive property? Or do you prefer collecting like terms. – Mike Dec 18 '11 at 1:20

When simplifying radicals the first step is to expose multiplicative dependencies by normalizing the radicands to be squarefree, i.e. pull out square factors. In your example we have $\rm 200 = 2\cdot 10^2\$ and $\ 32 = 2\cdot 4^2\$ so we obtain $\rm \sqrt{200}-\sqrt{32}\ = \sqrt{2\cdot 10^2}-\sqrt{2\cdot 4^2}\ =\ 10\ \sqrt{2} - 4\ \sqrt{2}\ =\ 6\ \sqrt{2}\:.$
@Jordan Yes, or, more generally, an $\rm\:n$'th root for integer (or rational) $\rm\:n\:.$ – Bill Dubuque Dec 17 '11 at 20:13
$$\sqrt{200}-\sqrt{32} = \sqrt{2\cdot 100}-\sqrt{2\cdot16} = \sqrt{2}\sqrt{100}-\sqrt{2}\sqrt{16} = \sqrt{2}(10-4) = 6\sqrt{2} = \sqrt{2\cdot36}=\sqrt{72}$$