How do I simplify this?  Anyone who knows and can give step-by-step answers, I would really appreciate it.  Thanks.

gibsonj338  Apr 21, 2018

hint: apply radical rule!


I'll post a solution later, but my answer is: \(\frac{\sqrt{2h-2x}}{h}-i \frac{\sqrt2\sqrt x}{h}\)

tertre  Apr 21, 2018



\(\frac{{\sqrt{-2(h+x)}-\sqrt{-2x}} }{h}    \\ \frac{{\sqrt{-1}\sqrt{2(h+x)}-\sqrt{-1}\sqrt{2x}} }{h}    \\ \frac{\sqrt{-1}\left[\sqrt{2(h+x)}-\sqrt{2x}\right]}{h}    \\ \frac{\left[\sqrt{2h+2x}-\sqrt{2x}\;\right]\;i}{h}    \\ or\\ \frac{\left[\sqrt{h+x}-\sqrt{x}\;\right]\sqrt2\;i}{h}    \\ \)


the final answer really depends on what you are doing it for.

Melody  Apr 21, 2018

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