Question

if x=√3+√2/√3-√2 and y=√3-√2/√3+√2 find x^2+y^2-xy

Answer 1

$\bold \red{x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \: and \: y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}$

$\bold \pink{We \: have \: to \: find \: x^2+xy+y^2}$

$\bold \green{First \: we \: calculate \: x^2}$

$\bold \blue{Consider \: x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}$

$\small \bold \green{We \: first \: rationalize \: the \: denominator \: by \: multiply}$

$\bold \red{and \: divide \: by {\sqrt{3}+\sqrt{2}}}$

We get,

$\bold \pink{x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} x}$

Simplify, we get,

$\begin{gathered} x=\frac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}\\\\ x=\frac{(\sqrt{3}+\sqrt{2})^2}{3-2}\\\\ x=(\sqrt{3}+\sqrt{2})^2\end{gathered}$

Thus, squaring both side we get,

$\bold \blue{x^2=((\sqrt{3}+\sqrt{2})^2)^2}$

using algebraic identity

$\bold \pink{(a+b)^2=a^2+b^2+2 ab}$

we have,

$\bold \green{x^2=(3+2+2\sqrt{6})^2=(5+2\sqrt{6})^2}$

$\bold \blue{Similarly, for \: y^2}$

$\bold \red{Consider y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}$

$\small \bold \green{We \: first \: rationalize \: the \: denominator \: by \: multiply}$

$\bold \green{and \: divide \: by \: {\sqrt{3}-\sqrt{2}} }$

we get,

$\bold \pink{x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}}$

Simplify, we get,

$\bold \red{y=(\sqrt{3}-\sqrt{2})^2}$

$\bold \pink{y^2=((\sqrt{3}-\sqrt{2})^2)^2}$

we have,

$y^2=(3+2-2\sqrt{6})^2=(5-2\sqrt{6})^2$

$then \: x^2+xy+y^2=(5+2\sqrt{6})^2+(5+2\sqrt{6})(5-2\sqrt{6})+(5-2\sqrt{6})^2$

Simplify , we get,

$\begin{gathered}x^2+xy+y^2=25+24+20\sqrt{6}+25-24+25+24-20\sqrt{6}\\\\ x^2+xy+y^2=25+24+25+25\\\\ x^2+xy+y^2=99\end{gathered}$

$Thus, \: x^2+xy+y^2=99$