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CAT 2025Slot 2QAQues & Sol

AlgebraMedium

Question

The equations $3x^{2}-5x+p=0$ and $2x^{2}-2x+q=0$ have one common root. The sum of the other roots of this equations is

Options

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.

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: $\frac{8}{3} - 2r = \frac{8}{3} - 2 \cdot \frac{2p - 3q}{4}$ $= \frac{8}{3} - \frac{2p - 3q}{2}$ $= \frac{8}{3} - p + \frac{3}{2}q$

3. Solution

Answer = Option A ✅

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