How do you simplify 4/(sqrt2-5sqrt3)?

Aug 3, 2017

$\frac{4}{\sqrt{2} - 3 \sqrt{5}} = - \frac{4}{43} \left(\sqrt{2} + 3 \sqrt{5}\right) = - \frac{4}{43} \sqrt{2} - \frac{12}{43} \sqrt{5}$

Explanation:

The difference of squares identity can be written:

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

Using this with $a = \sqrt{2}$ and $b = 5 \sqrt{3}$, we can rationalise the denominator of the given expression by multiplying both numerator and denominator by $\sqrt{2} + 5 \sqrt{5}$...

$\frac{4}{\sqrt{2} - 3 \sqrt{5}} = \frac{4 \left(\sqrt{2} + 3 \sqrt{5}\right)}{\left(\sqrt{2} - 3 \sqrt{5}\right) \left(\sqrt{2} + 3 \sqrt{5}\right)}$

$\textcolor{w h i t e}{\frac{4}{\sqrt{2} - 3 \sqrt{5}}} = \frac{4 \left(\sqrt{2} + 3 \sqrt{5}\right)}{{\left(\sqrt{2}\right)}^{2} - {\left(3 \sqrt{5}\right)}^{2}}$

$\textcolor{w h i t e}{\frac{4}{\sqrt{2} - 3 \sqrt{5}}} = \frac{4 \left(\sqrt{2} + 3 \sqrt{5}\right)}{2 - 45}$

$\textcolor{w h i t e}{\frac{4}{\sqrt{2} - 3 \sqrt{5}}} = - \frac{4}{43} \left(\sqrt{2} + 3 \sqrt{5}\right)$

$\textcolor{w h i t e}{\frac{4}{\sqrt{2} - 3 \sqrt{5}}} = - \frac{4}{43} \sqrt{2} - \frac{12}{43} \sqrt{5}$