Question

If the roots of the equation $a{x}^2+bx+c=0$ are real and distinct. Then

• A. Both roots are greater than $-\frac{b}{2a}$
• B. Both roots are less than $-\frac{b}{2a}$
• C. One of the roots exceeds $-\frac{b}{2a}$
• D. None of the above

Answer 1

Answer: (C)

One of the roots exceeds $-\frac{b}{2a}$

Let $\alpha$ and $\beta$ be the roots of the equation $a{x}^2+bx+c=0$

The roots of the given equation are
$\alpha =\frac{-b-\sqrt{b^2-4ac}}{2a}$
and $\beta =\frac{-b+\sqrt{b^2-4ac}}{2a}$
Since, $\alpha ,\beta$ are real and distinct, therefore
$b^2-4ac>0$. Now, if $a>0$, then $\beta >-\frac{b}{2a}$ and if $a<0$, then $\alpha >-\frac{b}{2a}$.

Thus, one of the roots exceeds $-\frac{b}{2a}$.