Question 12

If $$a = \frac{\sqrt3 + \sqrt2}{\sqrt3 - \sqrt2}$$ and $$b = \frac{\sqrt3 - \sqrt2}{\sqrt3 + \sqrt2}$$, then what is the value of $$a^2 + b^2 - ab$$?

Solution

From given data , ab=1 .

And, $$a+b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{3+2+2\sqrt{6}+3+2-2\sqrt{6}}{3-2}=10\ .$$

So, $$a^2+b^2-ab=\left(a+b\right)^2-3ab=10^2-3.1=97\ .$$

A is correct choice.

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