Help proving an statement regarding an equation.

Question

Let there be an equation like this: $$x^3 +px+q = 0$$

I want to prove that it has one and only one solution $x$ when $p > 0$ This is suppose to be solved using calculus, so there could be Lagrange, Rolle theorems and also some use of derivatives. Any help would be appreciated.

Answer 1

If $p>0$, $\,f(x)=x^3+px+q$ is an increasing function since it is the sum of two increasing functions ($x^3$ and $px+q$). Since $\lim_{x\to \pm\infty}f(x)=\pm\infty$ and $f(x)$ is a continuous function, $f(x)$ has exactly one real root, and quite trivially such root is negative/equal to zero/positive depending on $q=f(0)$ being positive/equal to zero/negative.

Answer 2

Hint:

The derivative of $x^3 +px+q = 0$ is $3x^2 + p =0$ which has no root over real numbers.

More precisely, $3x^2 + p >0$ holds over all real numbers.