Question

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

• A. $90$
• B. $93$
• C. $92$
• D. $91$

Answer 1

The correct option is B $93$

$ab=1$

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

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

$=\frac{2\left(\right(\sqrt{3}{)}^2+\left(\sqrt{2}{)}^2\right)}{(\sqrt{3}{)}^2-(\sqrt{2}{)}^2}=10[\because (a+b{)}^2+(a-b{)}^2=2(a^2+b^2);(a+b\left)\right(a-b)=a^2-b^2]$

$a^2+b^2-5ab=(a+b{)}^2-7ab={10}^2-7=93$