CAT 2025

Slot 2

QA

Question & SolutionQues & Sol

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

1. Concept Used

• Topic: Exponential Equations + Substitution + Vieta's Formulas
• Formula: For a quadratic $ax^2 + bx + c = 0$, product of roots $= \frac{c}{a}$

2. Calculation

We start with the equation: $9^{x^2+2x-3} - 4(3^{x^2+2x-2}) + 27 = 0$

Step 1: Substitute $t = x^2 + 2x - 3$

Note that $x^2 + 2x - 2 = t + 1$, so the equation becomes: $9^t - 4 \cdot 3^{t+1} + 27 = 0$

Rewriting $9^t = 3^{2t}$ and $3^{t+1} = 3 \cdot 3^t$: $3^{2t} - 12 \cdot 3^t + 27 = 0$

Step 2: Substitute $y = 3^t$

The equation becomes a quadratic: $y^2 - 12y + 27 = 0$

Factoring: $(y - 9)(y - 3) = 0 \Rightarrow y = 9 \text{ or } y = 3$

Step 3: Back-substitute $y = 3^t$

• If $y = 3 \Rightarrow 3^t = 3^1 \Rightarrow t = 1$
• If $y = 9 \Rightarrow 3^t = 3^2 \Rightarrow t = 2$

Step 4: Solve for x using $t = x^2 + 2x - 3$

Case 1: t = 1 $x^2 + 2x - 3 = 1 \Rightarrow x^2 + 2x - 4 = 0$

Discriminant $= 4 + 16 = 20 > 0$ (two real roots exist)

By Vieta's formulas, product of roots $= \frac{-4}{1} = -4$

Case 2: t = 2 $x^2 + 2x - 3 = 2 \Rightarrow x^2 + 2x - 5 = 0$

Discriminant $= 4 + 20 = 24 > 0$ (two real roots exist)

By Vieta's formulas, product of roots $= \frac{-5}{1} = -5$

Step 5: Product of ALL possible values of x

The four roots come from two separate quadratics. The product of all four roots is: $(-4) \times (-5) = 20$

3. Solution

Answer = Option B (20) ✅

The product of all possible values of x is 20.