Roots of a Cubic Equation depending on the value of $a$

Question

Here is the question I am currently working on:

Show that if $a$ is real, then the equation $x^3+ax+2=0$ has three real roots if and only if $a\leq-3$

I know that if the discriminant, call it $\Delta$, of a cubic equation of the form $x^3+ax^2+bx+c=0$ is positive, then the cubic equation must have three distinct real roots. I am planning to split this problem, into $3$ separate cases, one where $a<-3$, $a=-3$, and $a>-3$. I'm not entirely sure if this is even the right approach or am I just making things more complicated than what it should seem.

Answer 1

The discriminant of a cubic of the form $x^3+px+q=0$ is $\Delta=-4p^3-27q^2$.

So the discriminant of your equation is: $ \Delta=-4a^3-4\cdot 27 $ and the equation has three real solution iff $$ -4(a^3+27)\ge0 \iff a^3\le -27 \iff a \le -3 $$ (for $a=-3$ two solutions are coincident).