Question

If $\alpha ,\beta$ are the roots of the equation $a{x}^2+bx+c=0$, then the value of $\frac{(a{\alpha }^2+c)}{(a\alpha +b)}+\frac{(a{\beta }^2+c)}{(a\beta +b)}$ is

• A. $\frac{b(b^2-2ac)}{4a}$
• B. $\frac{b^2-4ac}{2a}$
• C. $\frac{b(b^2-2ac)}{a^{2}c}$
• D. none of these

Answer 1

Answer: (C)

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

$a{\alpha }^2+c$
$=a({\alpha }^2+\frac{c}{a})$
$=a({\alpha }^2+\alpha .\beta )$
$=a\alpha (\alpha +\beta )$
$=a\alpha \left(\frac{-b}{a}\right)$
$=-b\alpha$.
$a\alpha +b$
$=a(\alpha +\frac{b}{a})$
$=a(\alpha -\alpha -\beta )$
$=-a\beta$
Similarly
$a{\beta }^2+c$
$=-b\beta$
and $a\beta +b$
$=-a\alpha$
Taking the ratio, we get
$\frac{b}{a}(\frac{\alpha }{\beta }+\frac{\beta }{\alpha })$
$=\frac{b}{a}\left(\frac{{\beta }^2+{\alpha }^2}{\alpha .\beta }\right)$
$=\frac{b}{a}\left(\frac{b^2-2ac}{ac}\right)$
$=\frac{b(b^2-2ac)}{a^{2}c}$