Question

If $\alpha$ and $\beta ,(\alpha <\beta )$ are two different roots of the equation $p{x}^2+qx+r=0,$ then

• A. $\alpha >-\frac{q}{2p}$
• B. $\alpha <-\frac{q}{2p}<\beta$
• C. $-\frac{q}{2p}>\beta$
• D. None of the above

Answer 1

The correct option is B $\alpha <-\frac{q}{2p}<\beta$

Let $f\left(x\right)=p{x}^2+qx+r$
Clearly, $f\left(\alpha \right)=f\left(\beta \right)=0$
Also $f$ is continuous in $[\alpha ,\beta ]$ and differentiable in $(\alpha ,\beta )$
$\therefore$ From, Rolle's theorem $f^{\prime }\left(x\right)=0$ for atleast one value in $(\alpha ,\beta )$
$\Rightarrow 2px+q=0$
$\Rightarrow x=-\frac{q}{2p}\in (\alpha ,\beta )$