Question

Let $x=\sqrt{3}+\sqrt{2}$ and $y=\frac{1}{\sqrt{3}+\sqrt{2}}$ then find the value of $(x+y{)}^2+(x-y{)}^2$.

Answer 1

Given, $x=\sqrt{3}+\sqrt{2}$......(1) and $y=\frac{1}{\sqrt{3}+\sqrt{2}}$ or, $y=\frac{\sqrt{3}-\sqrt{2}}{3-2}=\sqrt{3}-\sqrt{2}$.....(2).

From (1) and (2) we've,

$x+y=2\sqrt{3}$ and $x-y=2\sqrt{2}$.

Now,

$(x+y{)}^2+(x-y{)}^2$

$=4(3+2)$

$=20$.