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 =

• A. 9
• B. 5
• C. 17
• D. 7

Answer 1

The correct option is A

9

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

(Rationalising the denominator)

$=\frac{(\sqrt{5}+\sqrt{3}{)}^2}{(\sqrt{5}{)}^2-(\sqrt{3}{)}^2}-\frac{5+3+2\sqrt{5}\times \sqrt{3}}{5-3}=\frac{8+2\sqrt{15}}{2}=4+\sqrt{15}Similarly,y=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}=\frac{(\sqrt{5}-\sqrt{3}{)}^2}{(\sqrt{5}{)}^2-(\sqrt{3}{)}^2}=\frac{5+3-2\sqrt{5}\times \sqrt{3}}{5-3}=\frac{8-2\sqrt{15}}{2}=4-\sqrt{15}$

Now, $x+y+xy=4+\sqrt{15}+4-\sqrt{15}+(4+\sqrt{15})(4-\sqrt{15})$

$=8+\left[\right(4{)}^2-\left(\sqrt{15}{)}^2\right]=8+(16-15)$

$=8+1=9$