Question

If the equation $x^3+px+q=0$ has three real roots, then show that $4p^3+27q^2<0$

Answer 1

We have,

$P\left(x\right)=x^3+px+q$

If $P\left(x\right)=0$ has three real roots.

Then it must cut $x-axis$.

So,

$P\left(x\right)=x^3+px+q$

On differentiating and we get,

$P^{\prime }\left(x\right)=3x^2+p$

For maxima and minima,

$P^{\prime }\left(x\right)=0$

$3x^2+p=0$

$x^2=\frac{-p}{3}$

Since equation $P\left(x\right)=0$ has three real roots .

$x=\pm \sqrt{-\frac{p}{3}}$

Now,

$P^{\prime \prime }\left(x\right)=6x$

At $x=\sqrt{-\frac{p}{3}}$

$P^{\prime \prime }\left(x\right)=6\sqrt{-\frac{p}{3}}>0$

It is minima.

So,

$27q^2+4p^3<0$

Hence proved.