Question

If $\alpha ,\beta$ are the roots of the equation $x^2-(3+\sqrt{2^{5{\log }_{2}4}-{\log }_2({\log }_3({\log }_{6}216)})x-2(3^{{\log }_{3}2}-2^{{\log }_{2}3})=0$, then the equation whose roots are $-\alpha ,-\beta$ is

• A. $x^2+19x+2=0$
• B. $x^2+35x-2=0$
• C. $x^2+29x+2=0$
• D. $x^2+35x+2=0$

Answer 1

The correct option is D $x^2+35x+2=0$

$x^2-(3+\sqrt{2^{5{\log }_{2}4}-{\log }_2({\log }_3({\log }_{6}216)})x-2(3^{{\log }_{3}2}-2^{{\log }_{2}3})=0$
$\Rightarrow x^2-(3+\sqrt{2^{{\log }_2{4}^5}-{\log }_2({\log }_{3}3)})x+2=0$
$\Rightarrow x^2-(3+\sqrt{2^{10}})x+2=0$
$\Rightarrow x^2-(3+2^5)x+2=0$
$\Rightarrow x^2-35x+2=0$

Now, replace $x$ by $-x,$ we get
$x^2+35x+2=0$