Question

If $x=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}},y=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$, find the value of ${(x-y)}^2$.

• A. 60
• B. 48
• C. 35
• D. 30

Answer 1

Answer: (D)

30

$x+y=\frac{{(\sqrt{5}-\sqrt{3})}^2+{(\sqrt{5}+\sqrt{3})}^2}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2}$
$=\frac{2(5+3)}{5-3}=8$
$xy=1$
$\therefore {(x-y)}^2={(x+y)}^2-4xy=64-4=60$