Question

If $\alpha$ and $\beta$ are the roots of the equation $a{{x}^{2}}+$ $bx+c=0,\text{ }\alpha \beta =3$ and $a, b, c$ are in $A.P.$, then $\alpha +\beta$ is equal to

• A. $-4$
• B. 1
• C. 4
• D. $-\,2$

Answer 1

Answer: (D)

$-\,2$

Answer 2

Since, $\alpha$ and $\beta$ are the roots of the equation $a{{x}^{2}}+bx+c=0$
$\therefore$ $\alpha +\beta =-\frac{b}{a}$ and $\alpha \beta =\frac{c}{a}$
But $\alpha \beta =3$
$\therefore$ $3=\frac{c}{a}\Rightarrow c=3a$ ...(i)
Also a, b, c are in AP.
$\therefore$ $b=\frac{a+c}{2}$
$\Rightarrow$ $b=\frac{a+3a}{2}=2a$ Hence, $\alpha +\beta =-\frac{b}{a}=-\frac{2a}{a}=-2$