I have trouble doing this and I'm not sure why are they the same and what are the steps I need to do to reach the simplified answer.
For example ...
$\frac{-8}{\sqrt{128}}$
This is the same as
$\frac{-1}{\sqrt{2}} $
I'm not sure how to reach the simplified expression from the earlier expression .. Can anyone help me ? Thanks !
$\frac{-8}{\sqrt {128}} = $$\frac{-8}{\sqrt 2 \sqrt {64}} = \frac{(-1)(8)}{\sqrt2(8)} = \frac{-1}{\sqrt2}$
Here are the steps: $$\frac {-8}{\sqrt{128}}=$$Simplify the bottom fraction $$ \frac {-8}{8\sqrt{2}}$$ cancel out the 8's $$ \frac {-1}{\sqrt{2}}$$
$\frac{-8}{\sqrt{128}}=-(\frac{8}{\sqrt{128}})=-(\sqrt{\frac{8^2}{(\sqrt{128})^2}})=-(\sqrt{\frac{64}{128}})=-(\sqrt{\frac{1}{2}})=-(\frac{1}{\sqrt{2}})=\frac{-1}{\sqrt{2}}$
In general, put the minus symbol outside, then try to simplify the content of the square root of the perfect square of numerator and denominator (the will be $a^2$ and $b$):
$$\frac{-a}{\sqrt{b}}=-(\frac{a}{\sqrt{b}})=-(\sqrt{(\frac{a}{\sqrt{b}})^2})=-(\sqrt{\frac{a^2}{(\sqrt{b})^2}})=-(\sqrt{\frac{a^2}{b}})$$