Question

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

Answer 1

$x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ [Rationalising the equation]

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

$=\frac{3-\sqrt{2}\sqrt{3}-\sqrt{2}\sqrt{3}+2}{3-2}$

$=5-2\sqrt{6}$

Similarly for y

$y=5-2\sqrt{6}$

$f\left(x\right)=x^2+xy-y^2$ [substituting the value]

$=(5-2\sqrt{6}{)}^2+(5-2\sqrt{6}\left)\right(5-2\sqrt{6})-(5-2\sqrt{6}{)}^2$

$\Rightarrow 5(5-2\sqrt{6})-2\sqrt{6}(5-2\sqrt{6})$

$\Rightarrow 25-10\sqrt{6}-10\sqrt{6}+24$

$\Rightarrow 49-20\sqrt{6}$