If $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ and $y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$, then find the value of $x^{2}+y^{2} ?$
Now,
$x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ [multiplying numerator and denominator by $\sqrt{3}+\sqrt{2}$ ]
$=\frac{(\sqrt{3}+\sqrt{2})^{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}$ [using identity, $(a+b)(a-b)=a^{2}-b^{2}$ ]
$=\frac{(\sqrt{3})^{2}+(\sqrt{2})^{2}+2 \cdot \sqrt{3} \cdot \sqrt{2}}{3-2}$ [using identity, $(a+b)^{2}=a^{2}+b^{2}+2 a b$ ]
$=\frac{3+2+2 \sqrt{6}}{1}=3+2+2 \sqrt{6}$
$\therefore \quad x=5+2 \sqrt{6}$ ...(i)
On squaring both sides, we get
$x^{2}=(5+2 \sqrt{6})^{2}$
$=(5)^{2}+(2 \sqrt{6})^{2}+2 \cdot 5 \cdot 2 \sqrt{6}$ [using identity, $(a+b)^{2}=a^{2}+b^{2}+2 a b$ ]
$\Rightarrow x^{2}=25+24+20 \sqrt{6}=49+20 \sqrt{6}$
$\Rightarrow x^{2}=49+20 \sqrt{6}$ ........(ii)
$\therefore \quad y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{1}{x}=\frac{1}{5+2 \sqrt{6}}$ [from Eq. (i)]
$=\frac{1}{5+2 \sqrt{6}} \times \frac{5-2 \sqrt{6}}{5-2 \sqrt{6}}$ [multiplying numerator and denominator by $5-2 \sqrt{6}$ ]
$=\frac{5-2 \sqrt{6}}{(5)^{2}-(2 \sqrt{6})^{2}}=\frac{5-2 \sqrt{6}}{25-24}=\frac{5-2 \sqrt{6}}{1}$ [using identity, $(a-b)(a+b)=a^{2}-b^{2}$ ]
On squaring both sides, we get
$y^{2}=(5-2 \sqrt{6})^{2}$
$\Rightarrow$ $y^{2}=(5)^{2}+(2 \sqrt{6})^{2}-2 \times 5 \times 2 \sqrt{6}$ [using identity, $(a-b)^{2}=a^{2}+b^{2}-2 a b$ ]
$\Rightarrow \quad y^{2}=25+24-20 \sqrt{6}$
$\Rightarrow \quad y^{2}=49-20 \sqrt{6}$ ...(iii)
On adding Eqs. (ii) and (iii), we get
$x^{2}+y^{2}=49+20 \sqrt{6}+49-20 \sqrt{6}=98$