Number of roots of a polynomial in respect to a variable [closed]

Given $p(x) = x^3 - 9x + \alpha $, I need to determine the number of real roots of the equation $p(x) = 0 $ in respect to $\alpha$ parameter.

I'm really stuck on this exercise and can't see much of a path to follow.

Solution 1:

As mentioned in the comments, the idea is to calculate the relative extremes as a first step: Let $f(x)=x^3-9x$, then $f'(x)=3x^2-9=3(x^2-3)$. So $f'(x)=0\Leftrightarrow x=\pm\sqrt 3$. Since $f''(x)=6x$, we have $f''(-\sqrt 3)<0$ (Then $f(-\sqrt 3)=6\sqrt 3$ is a relative maximum) and $f''(\sqrt 3)>0$ (Then $f(\sqrt 3)=-6\sqrt 3$ is a relative minimum). Now we see the graphic of $f$: Then let $root(\alpha)=$ roots of the function $f$

$$ root(\alpha)=\left\{ \begin{array}{ccl} 1&,&\alpha\in (-\infty,-6\sqrt 3)\cup (6\sqrt 3,+\infty)\\ 2&,&\alpha= \pm 6\sqrt 3\\ 3&,&\alpha\in (-6\sqrt 3,6\sqrt 3)\\ \end{array} \right\} $$