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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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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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