Question

If are the roots of the equation \[ 2x^3 \;-\; 5x^2 \;+\; 4x \;-\; 3 \;=\; 0, \] then \[ \sum \alpha\beta(\alpha+\beta) \;=\;? \]

• A. \(8\)
• B. \(4\)
• C. \(2\)
• D. \(\tfrac{1}{2}\)

Hint:

For a cubic \(ax^3 + bx^2 + cx + d=0\), always convert to sums/products of roots via \(\alpha+\beta+\gamma=-\tfrac{b}{a}\), etc.
- Use \(\alpha+\beta = (\alpha+\beta+\gamma) - \gamma\) to handle expressions like \(\alpha\beta(\alpha+\beta)\).

Answer 1

The Correct Option is D

Step 1: Recall relationships among roots.
For the cubic \(2x^3 - 5x^2 +4x -3=0,\) let \(\alpha,\beta,\gamma\) be its roots. Then: \[ \alpha + \beta + \gamma = -\frac{-5}{2} = \frac{5}{2},\quad \alpha\beta + \beta\gamma + \gamma\alpha = \frac{4}{2} = 2,\quad \alpha\beta\gamma = -\frac{-3}{2} = \frac{3}{2}. \] Step 2: Simplify the expression.
We want \[ \sum_{\text{cyc}} \alpha\beta\,(\alpha + \beta) \;=\; \alpha\beta(\alpha+\beta) \;+\; \beta\gamma(\beta+\gamma) \;+\; \gamma\alpha(\gamma+\alpha). \] But \(\alpha+\beta = (\alpha+\beta+\gamma) - \gamma = \frac{5}{2} - \gamma\). Hence \[ \alpha\beta(\alpha+\beta) = \alpha\beta\Bigl(\frac{5}{2} - \gamma\Bigr) = \frac{5}{2}\,\alpha\beta - \alpha\beta\gamma. \] Summing cyclically, \[ \sum \alpha\beta(\alpha+\beta) = \frac{5}{2}\,(\alpha\beta + \beta\gamma + \gamma\alpha) \;-\; \alpha\beta\gamma \sum_{\text{cyc}} 1 = \frac{5}{2}\cdot 2 \;-\; \bigl(\tfrac{3}{2}\bigr)\cdot 3. \] \[ = 5 \;-\; \frac{9}{2} = \frac{10 - 9}{2} = \frac{1}{2}. \] Hence the required sum is \(\boxed{\tfrac{1}{2}}\).