# How do you simplify (sqrt 6 - sqrt 5) / (sqrt 6 + sqrt 5)?

Oct 16, 2015

$11 - 2 \sqrt{30}$

#### Explanation:

Your goal here is to rationalize the denominator by multiplying it by its conjugate.

The conjugate of a binomial can be determined by changing the sign of the second term. In your case, you would have

$\sqrt{6} + \sqrt{5} \to {\underbrace{\sqrt{6} \textcolor{red}{-} \sqrt{5}}}_{\textcolor{b l u e}{\text{conjugate}}}$

So, multiply the fraction by $1 = \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} - \sqrt{5}}$ to get

(sqrt(6) - sqrt(5))/(sqrt(6) + sqrt(5)) * (sqrt(6) - sqrt(5))/(sqrt(6) - sqrt(5)) = ((sqrt(6) - sqrt(5))(sqrt(6) - sqrt(5)))/((sqrt(6) + sqrt(5))(sqrt(6) - sqrt(5))

The denominator takes the form

$\textcolor{b l u e}{\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}}$

and can thus be written as

$\left(\sqrt{6} + \sqrt{5}\right) \left(\sqrt{6} - \sqrt{5}\right) = {\left(\sqrt{6}\right)}^{2} - {\left(\sqrt{5}\right)}^{2} = 6 - 5 = 1$

The numerator takes the form

$\textcolor{b l u e}{{\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}}$

and can be written as

$\left(\sqrt{6} - \sqrt{5}\right) \left(\sqrt{6} - \sqrt{5}\right) = {\left(\sqrt{6}\right)}^{2} - 2 \sqrt{30} + {\left(\sqrt{5}\right)}^{2}$

$= 6 - 2 \sqrt{30} + 5$

$= 11 - 2 \sqrt{30}$

The expression becomes

$\frac{\left(\sqrt{6} - \sqrt{5}\right) \left(\sqrt{6} - \sqrt{5}\right)}{\left(\sqrt{6} + \sqrt{5}\right) \left(\sqrt{6} - \sqrt{5}\right)} = \frac{11 - 2 \sqrt{30}}{1} = \textcolor{g r e e n}{11 - 2 \sqrt{30}}$