Question

If $\alpha, \beta$ are the roots of the quadratic equation $x^2-\left(3+2^{\sqrt{\log _2 3}}-3^{\sqrt{\log _3 2}}\right) x-2\left(3^{\log _3 2}-2^{\log _2 3}\right)=0$, then the value of $\alpha^2+\alpha \beta+\beta^2$ is equal to

• A. 3
• B. 5
• C. 7
• D. 11

Answer 1

Answer: (C)

We know that $\left.2^{\sqrt{\log _2 3}}=2^{\left(\log _2 3\right.}\right)\left(\frac{1}{\sqrt{\log _2 3}}\right)=\left(2^{\log _2 3}\right) \frac{1}{\sqrt{\log _2 3}}=3^{\sqrt{\log _3 2}}$ (Using base changing formula)
$\therefore$ The given equation becomes $x ^2-3 x +2=0$
$\Rightarrow \alpha=1, \beta=2$
Hence $\alpha^2+\alpha \beta+\beta^2=7$