Number of roots of a Cubic Polynomial Let $f(x) = x^3 + px +q$ prove that if $(\frac{p}{2})^2+(\frac{q}{3})^3 > 0$ and $p <0 $ Then the equation has 1 real solution
Getting $f'(x)$ and setting it to $0$
$$x = \pm\sqrt{\frac{-p}{3}}$$
using the positive value and substituting it to $f(x)$, we get that its $y$ value is 
$$\frac{\sqrt{-27p^3}}{27} + p\frac{\sqrt{-3p}}{3} + q$$
The $y$ value of the inflection point $(q)$ must be larger than the difference between the $y$ value of the inflection point and the $y$ value of the critical point.
$$q > q-(\frac{\sqrt{-27p^3}}{27} + p\frac{\sqrt{-3p}}{3} + q)$$
and I get
$$(\frac{p}{4})^2+(\frac{q}{3})^3 > 0$$ 
Where did i go wrong?

Let $q\geq0$.
Thus, $$f\left(\sqrt{\frac{-p}{3}}\right)=\sqrt{\frac{-p^3}{27}}+p\sqrt{\frac{-p}{3}}+q=$$
$$=\sqrt{\frac{-p^3}{27}}-3\sqrt{\frac{-p^3}{27}}+q=2\left(\frac{q}{2}-\sqrt{\frac{-p^3}{27}}\right)=$$
$$=\frac{2\left(\left(\frac{q}{2}\right)^2+\left(\frac{p}{3}\right)^3\right)}{\frac{q}{2}+\sqrt{\frac{-p^3}{27}}}>0,$$
which says that our equation has an unique real root.
The case $q<0$ is a similar, but we need to work with a minimum point. 
