# How do you rationalize the denominator and simplify (12-2sqrt6)/(8-5sqrt2)?

Feb 12, 2017

See the explanation

#### Explanation:

$\textcolor{b l u e}{\text{Preamble}}$

There is a trick to this. Apart from anything else, mathematicians do not like 'roots' in the denominator. It is considered good practice to 'get rid of them'.

The trick: multiply by its self but change the sign in the middle (for the second variable) .

Why? Consider: $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$

Observe that everything ends up being squared. So any square roots disappear.
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$\textcolor{b l u e}{\text{Answering the question}}$

color(green)((12-2sqrt6)/(8-5sqrt2)color(red)(xx1)" "=" "color(green)((12-2sqrt6)/(8-5sqrt2)color(red)(xx(8+5sqrt(2))/(8+5sqrt(2)) )

$= \frac{\left(12 - 2 \sqrt{6}\right) \left(8 + 5 \sqrt{2}\right)}{{8}^{2} - {\left(5 \sqrt{2}\right)}^{2}}$

$= \frac{96 + 60 \sqrt{2} - 64 \sqrt{6} - 10 \sqrt{12}}{64 - 50}$
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Consider $10 \sqrt{12} \to 10 \sqrt{{2}^{2} \times 3} = 20 \sqrt{3}$
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$= \frac{96 + 60 \sqrt{2} - 64 \sqrt{6} - 20 \sqrt{3}}{14}$

$= \frac{48 + 30 \sqrt{2} - 32 \sqrt{6} - 5 \sqrt{12}}{7}$

$\textcolor{b l u e}{\text{I will leave the rest for you to do}}$