Question

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

Answer 1

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

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

Rationalising with $\sqrt{3}-\sqrt{2}$

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

$=\frac{3+2-2\sqrt{6}}{1}=5-2\sqrt{6}$

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

Rationalising with $\sqrt{3}+\sqrt{2}$

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

$=\frac{3+2+2\sqrt{6}}{1}=5\sqrt{2}\sqrt{6}$

Now,

$a^2+b^2-5ab$

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

$=(5^2+(2\sqrt{6}{)}^2-\mathrm{2.5.2}\sqrt{6})+((5{)}^2+(2\sqrt{6}{)}^2+\mathrm{2.5.2}\sqrt{6})-5(25-10\sqrt{6}+10\sqrt{6}-24)$

$=(25+24-20\sqrt{6})+(25+24+20\sqrt{6})-5(25-24)$

$=(49-20\sqrt{6}+49+20\sqrt{6}-5(1\left)\right)$

$=98-5$

$=93$

$\therefore a^2+b^2-5ab=93$.