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I'm trying to simplify a problem that I have the answer to but when I check it with Wolfram it seems a bit more simplified and I can't think of what steps are taking place to get that result.

Somehow this: $\frac{1}{2\sqrt{1-\frac{x^2}{4}}}$

Becomes this: $\frac{1}{\sqrt{4-x^2}}$

And I'm not sure how. I'm guessing its some simple Algebra operation that I'm totally forgetting right now. Any help breaking it down is much appreciated.

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3 Answers 3

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We have that $2=\sqrt{4}$. Therefore $$2\sqrt{1-\frac{x^2}{4}} = \sqrt{4}\sqrt{1-\frac{x^2}{4}} = \sqrt{4\left(1-\frac{x^2}{4}\right)}=\sqrt{4-x^2}.$$

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  • $\begingroup$ I think this one helped. So is it safe for me to assume that the process would be: combine 1 and x^2/4 to become 4x^2/4, then when you bring sqrt(4) in the 4's in 4x^2/4 cancel and you end up with 4-x^2? $\endgroup$
    – sara97
    Commented Nov 5, 2020 at 2:33
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$a\sqrt{b} = \sqrt{a^2b}$.

This moves the 2 inside the $\sqrt{}$ where it becomes $2^2 = 4$.

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  • $\begingroup$ Sorry I said I should have said I understood that the 4 is from the 2. I just couldn't figure out how the 4 in x^2/4 cancels. $\endgroup$
    – sara97
    Commented Nov 5, 2020 at 2:28
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We have $$\frac{1}{2\sqrt{1-\frac{x^2}{4}}}=\frac{1}{2\sqrt{\frac{1}{4}(4-x^2)}}$$ $$=\frac{1}{2\sqrt{\frac{1}{4}}\sqrt{4-x^2}}=\frac{1}{\sqrt{4-x^2}}.$$

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