Question

If $x=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ and $y=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$, then find $x+y+xy$
$\left[4\right]$

Answer 1

Given: $x=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ and $y=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$
Consider $x=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$

Rationallsing the denominator.

$x=\frac{(\sqrt{5}+\sqrt{3}{)}^2}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}$
$x=\frac{(\sqrt{5}{)}^2+2\sqrt{15}+(\sqrt{3}{)}^2}{(\sqrt{5}{)}^2-(\sqrt{3}{)}^2}$
$x=\frac{5+2\sqrt{15}+3}{5-3}$
$x=\frac{8+2\sqrt{15}}{2}$
$x=4+\sqrt{15}$
$\left[1.5\right]$
Consider, $y=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$
Rationalising the denominator,

$y=\frac{(\sqrt{5}-\sqrt{3}{)}^2}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}$
$y=\frac{(\sqrt{5}{)}^2-2\sqrt{15}+(\sqrt{3}{)}^2}{(\sqrt{5}{)}^2-(\sqrt{3}{)}^2}$
$y=\frac{5-2\sqrt{15}+3}{5-3}$
$y=\frac{8-2\sqrt{16}}{2}$
$y=4-\sqrt{15}$
$\left[1.5\right]$
Now, $x+y+xy$
$=4+\sqrt{15}+4-\sqrt{15}+(4+\sqrt{15})(4-\sqrt{15})$
$=8+4^2-(\sqrt{15}{)}^2$
$=8+16-15$
$=9$
$\left[1\right]$