Question

Rationalise the denominator $\frac{1}{3+\sqrt{5}-2\sqrt{2}}$

Answer 1

$Given$ $\frac{1}{3+\sqrt{5}-2\sqrt{2}}$

$=$$\frac{1}{3+(\sqrt{5}-2\sqrt{2})}$ $\times$ $\frac{3-(\sqrt{5}-2\sqrt{2})}{3-(\sqrt{5}-2\sqrt{2})}$

$=$ $\frac{3-(\sqrt{5}-2\sqrt{2})}{3^2-(\sqrt{5}-2\sqrt{2}{)}^2}$

$=$ $\frac{3-(\sqrt{5}-2\sqrt{2})}{9-(5+8-4\sqrt{10})}$

$=$ $\frac{3-(\sqrt{5}-2\sqrt{2})}{-4+4\sqrt{10}}$

$=$ $\frac{3-(\sqrt{5}-2\sqrt{2})}{4(\sqrt{10}-1)}$$\times$ $\frac{\sqrt{10}+1}{\sqrt{10}+1}$

$=$ $\frac{3\sqrt{10}-\sqrt{50}+2\sqrt{20}+3-\sqrt{5}+2\sqrt{2}}{4(10-1)}$

$=$ $\frac{3\sqrt{10}-3\sqrt{2}+3\sqrt{5}+3}{4\left(9\right)}$

$=$ $\frac{\sqrt{10}-\sqrt{2}+\sqrt{5}+1}{4\left(3\right)}$

$=$ $\frac{\sqrt{10}-\sqrt{2}+\sqrt{5}+1}{12}$