Question

If $\alpha ,\beta$ are the roots of the quadratic equation $a{x}^2+bx+c=0$, then $\alpha {\beta }^2+{\alpha }^2\beta +\alpha \beta$ equals:

• A. $-\frac{bc}{a^2}$
• B. $0$
• C. $abc$
• D. $\frac{c\left(a-b\right)}{a^2}$

Answer 1

The correct option is D

$\frac{c\left(a-b\right)}{a^2}$

Explanation for the correct option:

Find the value of the expression:

Since, $\alpha ,\beta$ are the roots of the quadratic equation $a{x}^2+bx+c=0$,then

Sum of root $=-\frac{\mathrm{coefficient}\mathrm{of}x}{\mathrm{coefficient}\mathrm{of}{x}^2}$ and Product of root $=\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coefficient}\mathrm{of}{x}^2}$

Then,

$\alpha +\beta =\frac{-b}{a}...\left(1\right)$ and $\alpha ·\beta =\frac{c}{a}...\left(2\right)$

Now,

$\begin{array}{rcl}\alpha {\beta }^2+{\alpha }^2\beta +\alpha \beta & =& \alpha \beta \left(\alpha +\beta \right)+\alpha \beta \\ & =& \frac{c}{a}·\left(-\frac{b}{a}\right)+\frac{c}{a}\left[\mathrm{from}\left(1\right)\mathrm{and}\left(2\right)\right]\\ & =& -\frac{cb}{a^2}+\frac{c}{a}\\ & =& \frac{-bc+ca}{a^2}\\ & =& \frac{c\left(a-b\right)}{a^2}\end{array}$

Hence, the correct option is D.