CAT 2025

Slot 2

QA

Question & Solution

1. Concept Used

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

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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: $$(-4) \times (-5) = 20$$

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3. Solution

Answer = Option B (20) ✅

The product of all possible values of x is 20.