# Algebra: removing the denominator when variable is positive.

Can someone explain the steps to get from

$\frac{1}{[\frac{1}{2}x(1 - \sqrt{1-4/x}) -1]}$

to

$\frac{1}{2}x + \frac{1}{2}\sqrt{x(x - 4)} - 1$

when assuming x is positive?

I understand the simplification of the square root, but how did we get rid of the fraction form?

https://www.wolframalpha.com/input/?i=1%2F%5B.5x(1+-+sqrt(1-4%2Fx))+-1%5D

• Those two aren't equal. – Andrew Li Feb 23 '18 at 2:31
• It's when we assume x is positive, according to wolframalpha (see link) – algorithmatic Feb 23 '18 at 3:09
• But those two aren't equal... plus the expression in your question doesn't match the one in WA – Andrew Li Feb 23 '18 at 3:10
• Sorry I updated it. – algorithmatic Feb 23 '18 at 3:16

Hint:  $\,x\sqrt{1-4/x}=\sqrt{x(x-4)}\,$ when $\,x \ge 0\,$ , then using $\,(a-b)(a+b)=a^2-b^2\,$:
\begin{align} \left(\frac{x}{2} - \frac{\sqrt{x(x-4)}}{2} -1\right) \left(\frac{x}{2} + \frac{\sqrt{x(x-4)}}{2} -1\right) = \left(\frac{x}{2}-1\right)^2 - \left(\frac{\sqrt{x(x-4)}}{2}\right)^2 = \ldots \end{align}
• but how did you get to the first term ($\frac{x}{2} - \frac{\sqrt(x(x-4)}{2} - 1)$? – algorithmatic Feb 27 '18 at 0:11
• @algorithmatic That's the denominator $\frac{1}{2}x(1 - \sqrt{1-4/x}) -1$ $= \frac{x}{2}-\frac{x\sqrt{1-4/x}}{2}-1$ $=\frac{x}{2}-\frac{\sqrt{x^2(1-4/x)}}{2}-1$ $=\frac{x}{2}-\frac{\sqrt{x(x-4)}}{2}-1$. – dxiv Feb 27 '18 at 0:16
• Yes but how is the whole fraction flipped and turned into $\frac{1}{2}x + \frac{1}{2}\sqrt{x(x - 4)} - 1$? Are you saying I should work on the denominator separately, and then plug into the fraction? – algorithmatic Feb 27 '18 at 0:20
• @algorithmatic $\,\frac{1}{a}=b \iff ab=1\,$. You asked the former, my answer proves the latter. – dxiv Feb 27 '18 at 0:36