Question

If $x=\frac{\sqrt{3}+2}{\sqrt{3}-2},y=\frac{\sqrt{3}-2}{\sqrt{3}+2}$ .Find $x^2+y^2$

Answer 1

Consider the given value of x and y ,

$x=\frac{\sqrt{3}+2}{\sqrt{3}-2},y=\frac{\sqrt{3}-2}{\sqrt{3}+2}$

We have to find,

$x^2+y^2={\left(\frac{\sqrt{3}+2}{\sqrt{3}-2}\right)}^2+{\left(\frac{\sqrt{3}-2}{\sqrt{3}+2}\right)}^2$

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

$=\frac{{\sqrt{3}}^2+4+4\sqrt{3}+{\sqrt{3}}^2+4-4\sqrt{3}}{{\left[(\sqrt{3}-2)(\sqrt{3}+2)\right]}^2}=\frac{14}{{[{\sqrt{3}}^2-2^2]}^2}$

$=\frac{14}{{(-1)}^2}=14$

This is answer.