Question

If \(a\), \(b\), and \(c\) are the roots of \( 2x^3 - 3x^2 + px - 1 = 0 \) and the sum of two roots is 1, the value of \(p\) is:

• A. \(2\)
• B. \(1\)
• C. \(3\)
• D. \( \frac{1}{3} \)

Hint:

When working with cubic equations, use Vieta’s formulas to relate the sums and products of the roots to the coefficients. This can help simplify the process of solving for unknowns such as \(p\).

Answer 1

The Correct Option is C

Given the cubic equation: \[ 2x^3 - 3x^2 + px - 1 = 0 \] By Vieta’s formulas, we know the relationships between the roots \(a\), \(b\), and \(c\) for the cubic equation \(Ax^3 + Bx^2 + Cx + D = 0\): \[ a + b + c = -\frac{B}{A}, \quad ab + bc + ca = \frac{C}{A}, \quad abc = -\frac{D}{A}. \] For the equation \( 2x^3 - 3x^2 + px - 1 = 0 \), we have: \[ a + b + c = \frac{3}{2}, \quad ab + bc + ca = \frac{p}{2}, \quad abc = \frac{1}{2}. \] We are also given that the sum of two roots is 1, i.e., \(a + b = 1\). Therefore, \[ a + b + c = 1 + c = \frac{3}{2}, \] which implies that \[ c = \frac{1}{2}. \] Now, substitute \(c = \frac{1}{2}\) into the equation for \(ab + bc + ca\): \[ ab + b \left( \frac{1}{2} \right) + a \left( \frac{1}{2} \right) = \frac{p}{2}, \] \[ ab + \frac{a + b}{2} = \frac{p}{2}. \] Since \(a + b = 1\), we have: \[ ab + \frac{1}{2} = \frac{p}{2}. \] Thus, \[ ab = \frac{p}{2} - \frac{1}{2}. \] Now, we use the fact that \(abc = \frac{1}{2}\). Since \(c = \frac{1}{2}\), we have: \[ ab \cdot \frac{1}{2} = \frac{1}{2}, \] which gives: \[ ab = 1. \] Substitute \(ab = 1\) into the earlier equation \(ab = \frac{p}{2} - \frac{1}{2}\), we get: \[ 1 = \frac{p}{2} - \frac{1}{2}. \] Solving for \(p\), we get: \[ 1 + \frac{1}{2} = \frac{p}{2} \quad \Rightarrow \quad \frac{3}{2} = \frac{p}{2} \quad \Rightarrow \quad p = 3. \] Thus, the value of \(p\) is \(3\).