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 these two equations is:

• A. $\dfrac{5}{3} - p + q$
• B. $\dfrac{8}{3} + p - q$
• C. $\dfrac{8}{3} - p + \dfrac{3}{2}q$
• D. $p + q - 1$

Hint:

When two quadratics share a common root, equate the root expressions by eliminating the squared term. Using Vieta’s formulas then makes it easy to compute required expressions involving the other roots.

Answer 1

The Correct Option is C

Approach: The two "other" roots are found from sum-of-roots (Vieta), but they still carry the unknown common root \(\alpha\). Kill \(\alpha\) by combining the two equations so the \(\alpha^2\) term cancels, leaving \(\alpha\) in terms of \(p\) and \(q\).

Let \(\alpha\) be the common root. Let the first equation \(3x^2 - 5x + p = 0\) have roots \(\alpha,\beta\) and the second \(2x^2 - 2x + q = 0\) have roots \(\alpha,\gamma\).

Step 1: By Vieta, \[ \alpha + \beta = \frac{5}{3}, \qquad \alpha + \gamma = \frac{2}{2} = 1. \] Adding, \[ \beta + \gamma = \frac{5}{3} + 1 - 2\alpha = \frac{8}{3} - 2\alpha. \]

Step 2: Find \(\alpha\). Since \(\alpha\) satisfies both equations: \[ 3\alpha^2 - 5\alpha + p = 0, \qquad 2\alpha^2 - 2\alpha + q = 0. \] Multiply the first by \(2\) and the second by \(3\) to match the \(\alpha^2\) terms: \[ 6\alpha^2 - 10\alpha + 2p = 0, \qquad 6\alpha^2 - 6\alpha + 3q = 0. \] Subtract: \[ -4\alpha + 2p - 3q = 0 \Rightarrow \alpha = \frac{2p - 3q}{4}. \]

Step 3: Substitute \(2\alpha = \dfrac{2p-3q}{2} = p - \dfrac{3q}{2}\): \[ \beta + \gamma = \frac{8}{3} - \left(p - \frac{3q}{2}\right) = \frac{8}{3} - p + \frac{3}{2}q. \]

Final answer: \( \dfrac{8}{3} - p + \dfrac{3}{2}q \) — option 3.

Answer 2

Let the common root be $\alpha$. Let the other roots of the first and second equations be $\beta$ and $\gamma$ respectively. The equations are: 1. \(3x^2 - 5x + p = 0\) with roots $\alpha$, $\beta$ 2. \(2x^2 - 2x + q = 0\) with roots $\alpha$, $\gamma$ We want the sum of the other roots: \[ \beta + \gamma. \] Step 1: Use the sum of roots formula for each equation. For \(3x^2 - 5x + p = 0\): \[ \alpha + \beta = \frac{5}{3} \quad\Rightarrow\quad \beta = \frac{5}{3} - \alpha. \] For \(2x^2 - 2x + q = 0\): \[ \alpha + \gamma = 1 \quad\Rightarrow\quad \gamma = 1 - \alpha. \]
Step 2: Sum of the other roots. \[ \beta + \gamma = \left(\frac{5}{3} - \alpha\right) + (1 - \alpha) = \frac{8}{3} - 2\alpha. \tag{1} \]
Step 3: Use the fact that $\alpha$ is a common root. From the first equation: \[ 3\alpha^2 - 5\alpha + p = 0. \] From the second equation: \[ 2\alpha^2 - 2\alpha + q = 0. \] Multiply the first equation by 2: \[ 6\alpha^2 - 10\alpha + 2p = 0. \] Multiply the second equation by 3: \[ 6\alpha^2 - 6\alpha + 3q = 0. \] Subtract: \[ (6\alpha^2 - 10\alpha + 2p) - (6\alpha^2 - 6\alpha + 3q) = 0, \] \[ -4\alpha + 2p - 3q = 0, \] \[ 4\alpha = 2p - 3q, \] \[ \alpha = \frac{2p - 3q}{4}. \tag{2} \]
Step 4: Substitute (2) into (1). \[ \beta + \gamma = \frac{8}{3} - 2\left(\frac{2p - 3q}{4}\right) = \frac{8}{3} - \frac{2p - 3q}{2} = \frac{8}{3} - p + \frac{3}{2}q. \] Thus, the sum of the other roots is: \[ \boxed{\frac{8}{3} - p + \frac{3}{2}q}. \]