CAT 2025 Slot 2 · Quant · Question 5

Put t={x}^{2}+2x-3, so that the middle exponent {x}^{2}+2x-2 is t+1.

The equation becomes {9}^{t}-4\cdot{3}^{t+1}+27=0.

Since {9}^{t}={\left({3}^{t}\right)}^{2} and 4\cdot{3}^{t+1}=12\cdot{3}^{t}, substituting k={3}^{t} turns it into {k}^{2}-12k+27=0.

Factorising, \left( k-3 \right)\left( k-9 \right)=0,

so k=3 or k=9.

{3}^{t}=3 gives t=1, and {3}^{t}=9 gives t=2.

Returning to x: t=1 gives {x}^{2}+2x-4=0,

and t=2 gives {x}^{2}+2x-5=0.

Their discriminants are 20 and 24, both positive, so there really are four real values of x.

By Vieta the two products of roots are -4 and -5,

so the product of all four is \left( -4 \right)\times \left( -5 \right)=20.

Shortcut. Do not solve for x. Once the two quadratics appear, the product of all roots is just the product of their constant terms (each has leading coefficient 1),

so \left( -4 \right)\left( -5 \right)=20 without touching a square root.