Question

If $a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}andb=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}},$ find the value of $a^2+b^2-5ab$

Answer 1

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

$ab=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}\times \sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=1$

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

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

$=\frac{5-2\sqrt{6}+5+2\sqrt{6}}{3-2}=10$

Now, $(x+y{)}^2=x^2+y^2+2xy$

$\Rightarrow x^2+y^2=(x+y)-2xy$

Replace $x=a,y=b$

$a^2+b^2=(a+b{)}^2-2ab$

$=(10{)}^2-2\times 1$

$(x+y)=10$ and $xy=1$

$=100-2=98$

$a^2+b^2-5ab=98-5\times 1$

$=98-5$

$=93$