Showing a cubic equation has at most one real root or three real roots Suppose we have $x^3 + \alpha x + \beta = 0$
Problem:
the cubic equation has one real root if $\alpha > 0$ and three real roots if $4 \alpha^3 + 27 \beta^2 < 0 $.
Try:
Let $f(x) = x^3 + \alpha x + \beta$. One has that $f'(x) = 3 x^2 + \alpha $. We have critical points when
$$ 3x^2 + \alpha = 0 \iff x^2 = - \frac{ \alpha }{3} $$
Clearly, if $\alpha > 0$, then we dont have critical points and $f'(x) > 0$ thus always increasing. Since $\lim_{x \to \infty} f(x) = \infty$ and $\lim_{x \to -\infty} = - \infty$, then there must be some $c$ such that $f(c) = 0$, so we have one real root.
Now, if $\alpha < 0$, then we have critical points $x = \pm \sqrt{ -\alpha/3 } $. but, here I am stuck. Any help would be greatly appreaciated.
 A: $x=-\sqrt{-\alpha\over3}$ is the maximum, $x=\sqrt{-\alpha\over3}$ is the minimum. 
If we want three real roots it must be:
$$[(-t)^3-\alpha t+\beta][t^3+\alpha t+\beta]<0,$$
(where $t=\sqrt{-\alpha\over3}$). In fact we know that if $f(x)=x^3+\alpha x+\beta:$ $$\lim_{x\to\pm\infty}f(x)=\pm\infty,$$ $f'(x)>0$ if $x<-\sqrt{-\alpha\over3}$; $f'(x)<0$ if $-\sqrt{-\alpha\over3}<x<\sqrt{{-\alpha\over3}}$ and $f'(x)>0$ if $x>\sqrt{-\alpha\over3}$, so if the above inequality is true, then the maximum must be positive and the minimum negative and the function intersects the $x-axis$ at three points (formally you can apply the Bolzano theorem), i.e. we must have three real roots.
If you calculate the expression you get:
$$-t^6-\alpha t^4-\beta t^3-\alpha t^4-\alpha^2 t^2-\alpha\beta t+\beta t^3+\alpha \beta t+\beta^2<0$$
$$-t^6-2\alpha t^4-\alpha^2 t^2+\beta^2<0$$
$$-{\alpha^3\over27}-{2\over9}\alpha^3+{\alpha^3\over3}+\beta^2<0$$
$${4\over27}\alpha^3+\beta^2<0$$
as required.
A: The fundamental thereom of algebra describes the roots of polynomials. In this case , the possibilities for a cubic equation is 1 complex pair (and one real root) or (three real roots). This is a little hand wavy, but you could use it more formally
A: Consider Vieta's formula for cubic polynomials:
given $P(x) = ax^3+bx^2+cx+d$ and roots $x_1,x_2,x_3$ for $P(x) = 0$, we know
$$x_1+x_2+x_3 = -\frac{b}{a}$$
$$x_1x_2+x_1x_3+x_2x_3=\frac{c}{a}$$
$$x_1x_2x_3 = -\frac{d}{a}$$
Your polynomial is the special case with $a=1,b=0$
Now assume a solution exists, where two roots are real and the third is imaginary, w.l.o.g. $x_1$ and $x_2$ are real, but $x_3$ is not. Let's say $x_3 = x_{3r} + i \cdot x_{3i}$, with $x_1,x_2,x_{3r},x_{3i} \in \mathbb{R}, x_{3i} \neq 0$.
Now:
$$x_1 + x_2 + x_3 = -b / a = 0$$
$$\Leftrightarrow x_1 + x_2 + x_{3r} + i \cdot x_{3i} = 0 + 0 \cdot i$$
Just looking at the imaginary parts, we get
$$\Rightarrow x_{3i} = 0$$
Which is a contradiction, thus two real roots is impossible. Since you can find examples with $0,1$ and $3$ real roots, you're done.
And $\alpha, \beta$ can be complex numbers for this.
