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This is not something I do very often, so I'm a bit dicey on the rules. I just want to make sure that I understand things right...

$$-\frac{1}{2}\cdot \sqrt{\frac{2}{5}} = -\sqrt{\frac{1}{4}}\cdot \sqrt{\frac{2}{5}}=-\sqrt{\frac{1}{10}}$$

I think this would normally work without a negative sign being involved, but I have a feeling that this doesn't work as I think it does.

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That works fine as you wrote it. You have to be careful when there are negative numbers insider the square roots. Then the law allowing radicands to be separated no longer apply. –  Ron Gordon Jan 11 '13 at 23:12

3 Answers 3

up vote 2 down vote accepted

Your work is just fine: you've shown you know that $\dfrac 12 = \sqrt{\dfrac{1}{4}},\;\;$ and that for any $x,y\in \mathbb{R^+\cup \{0\}},\;\;\sqrt x \cdot \sqrt y=\sqrt{xy}$.

The negative sign outside of the radicand has no impact on your operations: since the operations between terms is strictly multiplication, we can operate (multiply) as if the positive terms are entirely contained within parentheses, all of which is then multiplied by $-1$:

$$-\frac{1}{2}\cdot \sqrt{\frac{2}{5}} =-\left(\frac{1}{2}\cdot \sqrt{\frac{2}{5}}\right) = -\left(\sqrt{\frac{1}{4}}\cdot \sqrt{\frac{2}{5}}\right)$$ $$= -\left(\sqrt{\frac{1\cdot2}{4\cdot 5}}\right)=-\left(\sqrt{\frac{1}{10}}\right) = -\sqrt{\frac{1}{10}}$$

Note: the parentheses are used for illustration only: to make explicit that your computation is indeed correct. But, in fact, parentheses are not necessary.


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What you wrote is correct because $\sqrt{\frac{1}{4}}=\frac{1}{2}$ and for any $x,y\in \mathbb{R^+_0}$, $\sqrt x \sqrt y=\sqrt{xy}$. The minus sign (which is just multiplying by $-1)$ has no influence on the computation because you followed this rule.

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That's correct, the general rule is that you can put factor in roots if you put the same powers on them.

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