I'm asked to simplify $\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} – 2^{1/2}$ and am provided with the solution $\frac{-5\sqrt{2}-6}{7}$
I have tried several approaches and failed. Here's one path I took:
(Will try to simplify the left hand side fraction part first and then deal with the $-2^{1/2}$ later)
$\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}}$
The root of 16 is 4 and the root of 8 could be written as $2\sqrt{2}$ thus:
$\frac{2\sqrt{2}-4}{4-\sqrt{2}}$
Not really sure where to go from here so I tried multiplying out the radical in the denominator:
$\frac{2\sqrt{2}-4}{4-\sqrt{2}}$ = $\frac{2\sqrt{2}-4}{4-\sqrt{2}} * \frac{4+\sqrt{2}}{4+\sqrt{2}}$ = $\frac{(2\sqrt{2}-4)(4+\sqrt{2})}{16-2}$ =
(I become less certain in my working here)
$\frac{8\sqrt{2}*2(\sqrt{2}^2)-16-4\sqrt{2}}{14}$ = $\frac{8\sqrt{2}*4-16-4\sqrt{2}}{14}$ = $\frac{32\sqrt{2}-16-4\sqrt{2}}{14}$ = $\frac{28\sqrt{2}-16}{14}$
Then add back the $-2^{1/2}$ which can also be written as $\sqrt{2}$
This is as far as I can get. I don't know if $\frac{28\sqrt{2}-16}{14}-\sqrt{2}$ is still correct or close to the solution. How can I arrive at $\frac{-5\sqrt{2}-6}{7}$?
Best Answer
$$\begin{align} \frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}}-\sqrt{2}&=\frac{2\sqrt{2}-4}{4-\sqrt{2}}-\sqrt{2}\\ &=\frac{2\sqrt{2}-4}{4-\sqrt{2}}-\frac{4\sqrt{2}-2}{4-\sqrt{2}}\\ &=\frac{-2\sqrt{2}-2}{4-\sqrt{2}}\\ &=\frac{-2\sqrt{2}-2}{4-\sqrt{2}}~\cdot~\frac{4+\sqrt{2}}{4+\sqrt{2}}\\ &=\frac{-10\sqrt{2}-12}{14}\\ &=\frac{-5\sqrt{2}-6}{7} \end{align}$$