Question

If 2 and 3 are the two roots of the equation \[ 2x^3 + mx^2 - 13x + n = 0, \] then the values of \(m, n\) are respectively

• A. \(-5, 30\)
• B. \(5, -30\)
• C. \(-5, -30\)
• D. \(5, 30\)

Hint:

When two roots of a cubic are given, find the third using sum/product rules, then compute coefficients.

Answer 1

The Correct Option is A

Concept: For cubic \(ax^3 + bx^2 + cx + d = 0\) with roots \(\alpha, \beta, \gamma\): \[ \alpha + \beta + \gamma = -\frac{b}{a},\quad \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a},\quad \alpha\beta\gamma = -\frac{d}{a}. \]

Step 1: Let roots be \(2, 3, r\). Given equation: \(2x^3 + mx^2 - 13x + n = 0\), so \(a=2, b=m, c=-13, d=n\). Sum of roots: \[ 2 + 3 + r = -\frac{m}{2} \quad \Rightarrow \quad 5 + r = -\frac{m}{2} \quad \cdots (1) \] Sum of products taken two at a time: \[ 2\cdot 3 + 3r + 2r = \frac{c}{a} = \frac{-13}{2} \] \[ 6 + 5r = -\frac{13}{2} \] Multiply by 2: \(12 + 10r = -13 \quad \Rightarrow \quad 10r = -25 \quad \Rightarrow \quad r = -\frac{5}{2}\).

Step 2: Find \(m\) using (1). \[ 5 + \left(-\frac{5}{2}\right) = -\frac{m}{2} \] \[ \frac{10}{2} - \frac{5}{2} = \frac{5}{2} = -\frac{m}{2} \quad \Rightarrow \quad m = -5. \]

Step 3: Find \(n\) using product of roots. \[ 2 \cdot 3 \cdot r = -\frac{n}{a} = -\frac{n}{2} \] \[ 6 \cdot \left(-\frac{5}{2}\right) = -15 = -\frac{n}{2} \] Multiply by \(-2\): \(30 = n\). Thus \(m = -5, n = 30\).