0
$\begingroup$

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.

$\endgroup$

3 Answers 3

0
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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
    Nov 5, 2020 at 2:33
0
$\begingroup$

$a\sqrt{b} = \sqrt{a^2b}$.

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

$\endgroup$
1
  • $\begingroup$ 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
    Nov 5, 2020 at 2:28
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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