Question

If $x=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$ and $y=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ , find the value of $x^2+y^2-6xy$.

Answer 1

Given that

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

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

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

$x=4-\sqrt{15}$

$x^2=31-8\sqrt{15}$ .....................................(1)

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

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

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

$y=4+\sqrt{15}$

$y^2=31+8\sqrt{15}$ .....................................(2)

$xy=[4-\sqrt{15}]\times [4+\sqrt{15}]$

$\Rightarrow xy=1$ ........................................(3)

Now we have to find out the value of given expression

$x^2+y^2-6xy=(31-8\sqrt{15})+(31+8\sqrt{15})-6$

$=62-6=56$

Hence, $56$ is the answer