What are the necessary and sufficient conditions for the cubic equation to have at least 1 positive real root?
I'm just dealing with the $2$ simplest cases.
Case-1:
$$x^3+px+q=0$$
where, $p<0$ and $q>0$
Case-2:
$$x^3+px+q=0$$
where, $p<0$ and $q≤0.$
My attempts:
I wrote $x=y-\frac{p}{3y}$,
Then, I get $y^6+qy^3-\frac{p^3}{27}=0$
If $x>0$, then $y>\frac{p}{3y}$.
If $p<0, q>0$ then $y<0$. This gives
$$y>\frac{p}{3y}\implies y^2<\frac p3<0$$
I got $y\not\in\mathbb R$.
So,if $y\not\in\mathbb R$, then the inequality $y>\frac{p}{3y}$ will not work. But I know that a cubic equation with real coefficients must always have at least one real root. But this information is not helpful.
Hint: The derivative of this polynomial is $3x^2+p$, which has either a positive and a negative root, or no root at all. Since the limit as $x\to +\infty$ of $x^3+px+q$ is $+\infty$, you want to know whether there exists an $x\in\mathbb R_+$ such that $x^3+px+q$ is negative.
Full answer: From what I've stated above, the minimum of $x^3+px+q$ on $\mathbb R_+$ is either at $x=0$ (when $p>0$) or at $x=\sqrt{-p/3}$ (when p<0).
If $p\ge0$, then this minimum is simply $q$. We can see that if $q\le0$, then there is indeed a positive real root. In fact this remains true if $p<0$: $q\le0$ is a sufficient condition.
If $p<0$, we therefore only consider the case $q>0$. Then the minimum is $$\sqrt{-p/3}^3 +p\sqrt{-p/3}+q = \frac{2p\sqrt{-p}}{3\sqrt{3}}+q,$$
which is negative if and only if $$-\frac{4p^3}{27}\ge q^2.$$
In summary, $x^3+px+q$ has a real positive root if and only if $q\le0$ or $-4p^3\ge27q^2$.