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
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is $\sqrt{2\times 100} - \sqrt{2\times 16}$ of any help? – Dec 17 '11 at 17:38
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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 $6*\sqrt{2}$ – Dec 17 '11 at 17:40
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You mean the distributive property? Or do you prefer collecting like terms. – Mike Dec 18 '11 at 01:20
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$$\sqrt{200}-\sqrt{32} = \sqrt{2\cdot 100}-\sqrt{2\cdot16} = \sqrt{2}\sqrt{100}-\sqrt{2}\sqrt{16} = \sqrt{2}(10-4) = 6\sqrt{2} = \sqrt{2\cdot36}=\sqrt{72}$$
Asaf Karagila
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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 $\rm \sqrt{200}-\sqrt{32}\ = \sqrt{2\cdot 10^2}-\sqrt{2\cdot 4^2}\ =\ 10\ \sqrt{2} - 4\ \sqrt{2}\ =\ 6\ \sqrt{2}\:.$
When you go on to study the Galois theory of radical extensions (Kummer theory) you will learn general results saying roughly that these are the only type of algebraic dependencies that can occur, so this simple-minded approach will work generally. For some general algorithms see my post here and see Bill Gosper's reply there for some striking radical identities (if anyone deserves to be called a modern equivalent of Ramanujan then Gosper is surely a strong candidate).
Bill Dubuque
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@Jordan Yes, or, more generally, an $\rm:n$'th root for integer (or rational) $\rm:n:.$ – Bill Dubuque Dec 17 '11 at 20:13