Question

If $x=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ and $y=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$, then $x+y+xy=$

Answer 1

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

Consider $x=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$

Rationalising the denominator,

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

$x=\frac{(\sqrt{5}{)}^2+2\sqrt{15}+(\sqrt{3}{)}^2}{(\sqrt{5}{)}^2-(\sqrt{3}{)}^2}$

$x=\frac{5+2\sqrt{15}+3}{5-3}$

$x=\frac{8+2\sqrt{15}}{2}$

$x=4+\sqrt{15}$

Consider $y=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

Rationalising the denominator,

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

$y=\frac{(\sqrt{5}{)}^2-2\sqrt{15}+(\sqrt{3}{)}^2}{(\sqrt{5}{)}^2-(\sqrt{3}{)}^2}$

$y=\frac{5-2\sqrt{15}+3}{5-3}$

$y=\frac{8-2\sqrt{15}}{2}$

$y=4-\sqrt{15}$

Now, $x+y+xy$

$=4+\sqrt{15}+4-\sqrt{15}+(4+\sqrt{15})(4-\sqrt{15})$

$=8+4^2-(\sqrt{15}{)}^2$

$=8+16-15$

$=9$

Hence, (a) is the correct option.