# How do you simplify (4+sqrt2 )div (8 - sqrt2)?

Sep 20, 2015

$\frac{17 + 6 \sqrt{2}}{31}$

#### Explanation:

Let's rewrite this in the fractional form:
$\frac{4 + \sqrt{2}}{8 - \sqrt{2}}$

Then, I would multiply, top and bottom, by $\left(8 + \sqrt{2}\right)$ so that:
$\frac{4 + \sqrt{2}}{8 - \sqrt{2}} \cdot \frac{8 + \sqrt{2}}{8 + \sqrt{2}}$

(Note that $\frac{8 + \sqrt{2}}{8 + \sqrt{2}} = 1$, so multiplying by this factor does not change the results at all).

I chose $\left(8 + \sqrt{2}\right)$ because
$\left(a - b\right) \cdot \left(a + b\right) = {a}^{2} - {b}^{2}$
and I am hoping it will make the solution easier.

Then the fraction becomes:
$\frac{\left(4 + \sqrt{2}\right) \left(8 + \sqrt{2}\right)}{{8}^{2} - {\left(\sqrt{2}\right)}^{2}} = \frac{32 + 12 \sqrt{2} + 2}{64 - 2}$
which is
$= \frac{34 + 12 \sqrt{2}}{62}$
We can factor out 2 from the top and bottom:
$= \frac{2 \cdot \left(17 + 6 \sqrt{2}\right)}{2 \cdot 31}$
$= \frac{17 + 6 \sqrt{2}}{31}$