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How do you do this?

Is it first simplifying $\sqrt{40}.$ If I do that I get $-5\sqrt 4\sqrt{10}$

Which gives me $-10\sqrt{10}.$

I don't think I am correct.

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  • $\begingroup$ I don't think I am correct. Why not? $\endgroup$ – dxiv Apr 18 '17 at 23:00
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You are correct. This is because of $(a\cdot b)^n=a^n\cdot b^n$. So, You have $-5\cdot (40)^{\frac{1}{2}}=-5\cdot (4\cdot10)^{\frac{1}{2}}=-5\cdot (4)^{\frac{1}{2}}\cdot(10)^{\frac{1}{2}}=-10\cdot 10^{\frac{1}{2}}=-10^{\frac{3}{2}}$

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Nope, you're exactly right. $-10 \sqrt{10}$ is correct.

$\sqrt {40} = \sqrt {4} \times \sqrt{10}$, and as $\sqrt{4} = 2$, then everything else snaps into place once you multiply by $-5$ and attach the surd.

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  • $\begingroup$ Thanks for cleaning up...I'm still new to MathJax so a few edits here and there help me out too. $\endgroup$ – bjcolby15 Apr 20 '17 at 0:40
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You are correct!

$-5\sqrt{40} = -5\sqrt{4\cdot 10}=-5\cdot\sqrt{2^2\cdot 10}=-5\cdot 2\cdot\sqrt{10}=-10\sqrt{10}$

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