if $x^3 + px^2+qx+r = 0$ has three real roots, show that $p^2 \ge 3q$

Question

If $x^3 + px^2+qx+r = 0$ has three real roots, show that $p^2 \ge 3q$

Can I get help on this problem, thanks in advance!

Answer 1

Hint: The existence of $3$ real roots of $P(x)=x^3+px^2+qx+r$ implies the existence of how many real roots of $P'(x)=3x^2+2px+q$?

Answer 2

Alternative hint: Express $p$ and $q$ in terms of the roots using Vieta's relations and see what transpires.

Answer 3

$f(x) = x^3 + px^2 +qx+r$
$f'(x) = 3x^2 + 2px +q$
Since $\operatorname{discr}{f(x)} > 0$
Therefore $\operatorname{discr}{f'(x)} > 0$
Thus
$\Rightarrow 4p^2 - 12q > 0$
$\Rightarrow p^2 > 3q$