Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
$\sqrt {100 \cdot 2}-\sqrt {16 \cdot 2}$ –  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
add comment

2 Answers 2

up vote 2 down vote accepted

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).

share|improve this answer
Is a radical a square root? –  user138246 Dec 17 '11 at 20:00
@Jordan Yes, or, more generally, an $\rm\:n$'th root for integer (or rational) $\rm\:n\:.$ –  Bill Dubuque Dec 17 '11 at 20:13
add comment

$$\sqrt{200}-\sqrt{32} = \sqrt{2\cdot 100}-\sqrt{2\cdot16} = \sqrt{2}\sqrt{100}-\sqrt{2}\sqrt{16} = \sqrt{2}(10-4) = 6\sqrt{2} = \sqrt{2\cdot36}=\sqrt{72}$$

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.