Question

If the equation $x^3+px+q=0(p<0)$ has three distinct real roots, then

• A. $4p^3+9q^2<0$
• B. $4p^3+27q^2<0$
• C. $2p^3+27q^2>0$
• D. $2p^2+27q^2<0$

Answer 1

The correct option is B $4p^3+27q^2<0$

$f\left(x\right)=x^3+px+q=0$
$f^{\prime }\left(x\right)=3x^2+p$
$f^{\prime \prime }\left(x\right)=6x$
$\therefore f\left(x\right)$ must have one maximum and one minimum.
$\therefore f^{\prime }\left(x\right)=0$
$\Rightarrow x=\pm \sqrt{\frac{-p}{3}}\{\because p<0\}$
where, $f$ is maximum at $x=-\sqrt{\frac{-p}{3}}\{\because f^{\prime \prime }(x)<0\}$
and minimum at $x=\sqrt{\frac{-p}{3}}\{\because f^{\prime \prime }(x)>0\}$
$f$ has three real roots so,
$f(-\sqrt{\frac{-p}{3}})>0$ and $f\left(\sqrt{\frac{-p}{3}}\right)<0$
Let $\sqrt{\frac{-p}{3}}=k$
$\Rightarrow f(-\sqrt{\frac{-p}{3}})\cdot f\left(\sqrt{\frac{-p}{3}}\right)<0$
$\Rightarrow (q+(k^3+pk))(q-(k^3+pk))<0$
$\Rightarrow q^2-{(k^3+pk)}^2<0$
$\Rightarrow q^2+\frac{4p^3}{27}<0$
$\Rightarrow 4p^3+27q^2<0$