# How do you rationalize the denominator and simplify (3+sqrt18) /( 1 + sqrt8 )?

Jul 15, 2016

$\frac{3}{7} \left(3 + \sqrt{2}\right) \leftarrow \text{ corrected solution}$

#### Explanation:

A very useful relationship to remember is ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
In the context of this question we can change
$1 + \sqrt{8} \text{ to } {1}^{2} - {\left(\sqrt{8}\right)}^{2}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Multiply by 1 but in the form of $1 = \frac{1 - \sqrt{8}}{1 - \sqrt{8}}$

" "color(red)("Error - should be "sqrt(18))
$\text{ } \textcolor{red}{\downarrow}$
$\cancel{\frac{\left(3 + \textcolor{red}{\sqrt{8}}\right) \left(1 - \sqrt{8}\right)}{\left(1 + \sqrt{8}\right) \left(1 - \sqrt{8}\right)}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{g r e e n}{\text{Corrected!!!}}$

$\frac{\left(3 + \sqrt{18}\right) \left(1 - \sqrt{8}\right)}{\left(1 + \sqrt{8}\right) \left(1 - \sqrt{8}\right)} = \frac{3 - 3 \sqrt{8} + \sqrt{18} - \sqrt{8 \times 18}}{1 - 8}$

$= \frac{3 - 3 \sqrt{2 \times {2}^{2}} + \sqrt{2 \times {3}^{2}} - \sqrt{2 \times {2}^{2} \times 2 \times {3}^{2}}}{- 7}$

$= \frac{3 - 6 \sqrt{2} + 3 \sqrt{2} - 12}{- 7}$

$= \frac{- 9 - 3 \sqrt{2}}{- 7} \text{ "=" } + \frac{3}{7} \left(3 + \sqrt{2}\right)$

Got there in the end! Amazing how much a typo can mess things up!

Jul 15, 2016

$\frac{3}{7} \left(3 + \sqrt{2}\right)$

#### Explanation:

Rationalisation factor of $a + \sqrt{b}$ is $a - \sqrt{b}$ and vice-versa.

Given Exp. $= \frac{3 + \sqrt{18}}{1 + \sqrt{8}} = \frac{3 + \sqrt{9 \cdot 2}}{1 + \sqrt{4 \cdot 2}}$

$= \frac{3 + 3 \sqrt{2}}{1 + 2 \sqrt{2}}$

Multiplying, in Nr. & Dr by the rationalisation factor of Dr., i.e., by$\left(1 - 2 \sqrt{2}\right)$, the Exp.

$= \frac{\left(3 + 3 \sqrt{2}\right) \left(1 - 2 \sqrt{2}\right)}{\left(1 + 2 \sqrt{2}\right) \left(1 - 2 \sqrt{2}\right)} ,$

$= \frac{3 + 3 \sqrt{2} - 6 \sqrt{2} - 6 \cdot 2}{{1}^{2} - {\left(2 \sqrt{2}\right)}^{2}} ,$

$= \frac{- 9 - 3 \sqrt{2}}{1 - 8} , = \frac{3}{7} \left(3 + \sqrt{2}\right) .$