# Prove the following

Question:

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 $x^{2}+y^{2} ?$

Solution:

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

$x^{2}+y^{2}=49+20 \sqrt{6}+49-20 \sqrt{6}=98$