CAT 2025 Slot 2 QA Question 18: If $9^{x^2 + 2x - 3} - 4(3^{x^2 + 2x - 2}) + 27 = 0$, then the product of all possible values of $x$ is

Question

If $9^{x^2 + 2x - 3} - 4(3^{x^2 + 2x - 2}) + 27 = 0$, then the product of all possible values of $x$ is

Accepted Answer

Let $t = x^2 + 2x - 3$. Then $3^{x^2 + 2 x - 2} = 3^{t+1}$.

$9^t - 4(3^{t+1}) + 27 = 0$
$3^{2t} - 12(3^t) + 27 = 0$

Let $y = 3^t$.
$y^2 - 12y + 27 = 0$
$(y-3)(y-9)=0$

Hence $y = 3$ or $y = 9$, so $t = 1$ or $t = 2$.

For $t = 1$:
$x^2 + 2x - 3 = 1$
$\Rightarrow x^2 + 2x - 4 = 0$
The product of it

If 9x2+2x−3−4(3x2+2x−2)+27=09^{x^2 + 2x - 3} - 4(3^{x^2 + 2x - 2}) + 27 = 09x2+2x−3−4(3x2+2x−2)+27=0, then the product of all possible values of xxx is