Question

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

Answer 1

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,$(a+b)(a-b)=a^2-b^2]$
$=\frac{(\sqrt{3}{)}^2+(\sqrt{2}{)}^2+2.\sqrt{3}.\sqrt{2}}{3-2}$ [Using identity, $(a+b{)}^2=a^2+b^2+2ab]$
$=\frac{3+2+2\sqrt{6}}{1}=3+2+2\sqrt{6}\therefore x=5+2\sqrt{6}-\left(i\right)$
On squaring both sides, we get,
$x^2=(5+2\sqrt{6}{)}^2$
$=(5{)}^2+(2\sqrt{6}{)}^2+\mathrm{2.5.2}\sqrt{6}$ [Using identity, $(a+b{)}^2=a^2+b^2+2ab]$
$\Rightarrow x^2=25+24+20\sqrt{6}=49+20\sqrt{6}\Rightarrow x^2=49+20\sqrt{6}-\left(ii\right)\therefore y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{1}{x}=\frac{1}{5+2\sqrt{6}}$
$=\frac{1}{5+2\sqrt{6}}\times \frac{5-2\sqrt{6}}{5-2\sqrt{6}}$ [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,$(a-b{)}^2=a^2+b^2-2ab]$
$\Rightarrow y^2=25+24-20\sqrt{6}\Rightarrow y^2=49-20\sqrt{6}\left(iii\right)$
On adding eqs. (ii) and (iii) we get,
$x^2+y^2=49+20\sqrt{6}+49-20\sqrt{6}=98$