# Question 3f390

Mar 16, 2017

$- \frac{2}{3} \sqrt{3} - \sqrt{2}$

#### Explanation:

Consider the just the denominator for a moment.

Demonstrating a principle by example:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Now compare to:

${\left(\sqrt{12}\right)}^{2} - {\left(\sqrt{18}\right)}^{2} = \left(\sqrt{12} - \sqrt{18}\right) \left(\sqrt{12} + \sqrt{18}\right)$

So to remove the roots from the denominator we need to 'force' it to be $\left(\sqrt{12} - \sqrt{18}\right) \left(\sqrt{12} + \sqrt{18}\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Getting rid of the roots in the denominator}}$

Multiply by 1 and you do not change the value. However, 1 comes in many forms so you can change the way numbers look without changing their intrinsic value.

$\textcolor{g r e e n}{\frac{2}{\sqrt{12} - \sqrt{18}} \textcolor{red}{\times 1}}$

color(green)(2/(sqrt(12)-sqrt(18))color(red)( xx(sqrt(12)+sqrt(18))/(sqrt(12+sqrt(18))#

$\frac{2 \left(\sqrt{12} + \sqrt{18}\right)}{{\left(\sqrt{12}\right)}^{2} - {\left(\sqrt{18}\right)}^{2}}$

$\frac{2 \left(\sqrt{12} + \sqrt{18}\right)}{12 - 18}$

$\frac{2 \left(\sqrt{12} + \sqrt{18}\right)}{- 6}$
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$\textcolor{b l u e}{\text{Simplifying}}$

$\frac{{\cancel{2}}^{1} \left(\sqrt{12} + \sqrt{18}\right)}{- {\cancel{6}}^{3}}$

$- \frac{1}{3} \left(\sqrt{12} + \sqrt{18}\right)$

But $\sqrt{12} = \sqrt{3 \times {2}^{2}} = 2 \sqrt{3}$
And $\sqrt{18} = \sqrt{2 \times {3}^{2}} = 3 \sqrt{2}$

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

$- \frac{2}{3} \sqrt{3} - \sqrt{2}$

Mar 16, 2017

$- \frac{2}{3} \sqrt{3} - \sqrt{2}$

#### Explanation:

Multiply the numerator/denominator of the fraction by the $\textcolor{b l u e}{\text{conjugate}}$ of the denominator.

The $\textcolor{b l u e}{\text{conjugate"" of "sqrt12-sqrt18" is }} \sqrt{12} \textcolor{red}{+} \sqrt{18}$

Considering the denominator.

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \left(\sqrt{a} \times \sqrt{a}\right) = a$

$\text{Expand } \left(\sqrt{12} - \sqrt{18}\right) \left(\sqrt{12} \textcolor{red}{+} \sqrt{18}\right)$ using the FOIL method.

$= 12 + \cancel{\sqrt{18} \times \sqrt{12}} \cancel{- \sqrt{18} \times \sqrt{12}} - 18$

$= - 6 \leftarrow \textcolor{red}{\text{ rational denominator}}$

$\Rightarrow \frac{2}{\sqrt{12} - \sqrt{18}}$

$= \frac{2}{\sqrt{12} - \sqrt{18}} \times \frac{\sqrt{12} + \sqrt{18}}{\sqrt{12} + \sqrt{18}}$

$= \frac{2 \left(\sqrt{12} + \sqrt{18}\right)}{- 6}$

$= - \frac{1}{3} \left(\sqrt{12} + \sqrt{18}\right)$

Simplifying $\sqrt{12} \text{ and } \sqrt{18}$

$\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2 \sqrt{3}$

$\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3 \sqrt{2}$

$\Rightarrow - \frac{1}{3} \left(\sqrt{12} + \sqrt{18}\right)$

$= - \frac{1}{3} \left(2 \sqrt{3} + 3 \sqrt{2}\right) \leftarrow \text{ this is acceptable}$

$= - \frac{2}{3} \sqrt{3} - \sqrt{2}$