CAT 2023

Slot 2

QA

Question & Solution

It is given that $2^{4x^{2}}-2^{2x^{2}+x+16}+2^{2x+30}=0$, which can be written as:

=>$\left(2^{2x^2}\right)^2-2^{2x^2}\cdot2^{x+15}\cdot2^1+\left(2^{x+15}\right)^{^2}=0$

=> $\left(2^{2x^2}-2^{x+15}\right)^{^2}=0$

=> $2^{2x^2}-2^{x+15}=0$ (Since $\left(a-b\right)^2\ =\ 0\ =>\ a-b\ =0$)

=> $2x^2\ =\ x+15$

=> $2x^2-x-15=0$ 

=> $2x^2-6x+5x-15=0$

=> $2x\left(x-3\right)+5\left(x-3\right)=0$

=> $\left(2x+5\right)\left(x-3\right)\ =\ 0$

Hence, the possible values of x are $-\frac{5}{2}$, and $3$, respectively.

Therefore, the sum of the possible values is $\left(3-\frac{5}{2}\right)=\frac{1}{2}$

The correct option is D