CAT 2025

Slot 2

QA

Question & Solution

1. Concept Used

• Topic: Quadratic Equations — Common Root & Vieta's Formulas
• Formula: For a quadratic $$ax^2 + bx + c = 0$$, sum of roots $$= -\frac{b}{a}$$. If $$r$$ is a common root of two quadratics, it satisfies both equations simultaneously, allowing elimination of $$r^2$$ to express $$r$$ in terms of coefficients.

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2. Calculation

Let $$r$$ be the common root shared by both equations.

By Vieta's formulas:

• Sum of roots of $$3x^2 - 5x + p = 0$$ is $$\frac{5}{3}$$
• Sum of roots of $$2x^2 - 2x + q = 0$$ is $$\frac{2}{2} = 1$$

If $$r$$ is the common root, the other root of the first equation is $$\frac{5}{3} - r$$, and the other root of the second equation is $$1 - r$$.

So, the required sum of the other two roots is: $$\left(\frac{5}{3} - r\right) + (1 - r) = \frac{8}{3} - 2r$$

Now we must express $$r$$ in terms of $$p$$ and $$q$$. Since $$r$$ is a common root, it satisfies both equations: $$3r^2 - 5r + p = 0 \quad \cdots (1)$$ $$2r^2 - 2r + q = 0 \quad \cdots (2)$$

To eliminate $$r^2$$, multiply equation (1) by 2 and equation (2) by 3: $$6r^2 - 10r + 2p = 0 \quad \cdots (3)$$ $$6r^2 - 6r + 3q = 0 \quad \cdots (4)$$

Subtract (3) from (4): $$(6r^2 - 6r + 3q) - (6r^2 - 10r + 2p) = 0$$ $$4r + 3q - 2p = 0$$ $$r = \frac{2p - 3q}{4}$$

Substitute back into the sum expression: $$\frac{8}{3} - 2r = \frac{8}{3} - 2 \cdot \frac{2p - 3q}{4}$$ $$= \frac{8}{3} - \frac{2p - 3q}{2}$$ $$= \frac{8}{3} - p + \frac{3}{2}q$$

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3. Solution

Answer = Option A ✅

The sum of the other two roots is $$\frac{8}{3} - p + \frac{3}{2}q$$.