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