Condition for fourth degree polynomial to have all real roots For what range of values of $a$ will the following fourth degree polynomial have all real roots:
$$x^4 - 2ax^2 + x + a^2 -a = 0$$
 A: Let denote $f(x)=x^4 - 2ax^2 + x + a^2 -a = 0$. We have
$$f'(x)=4x^3-4ax+1\quad;\quad f''(x)=12x^2-4a$$
If $f''\geq0$ then $f'$ is increasing and then $f$ has at most two real roots so a necessary condition to have 4 real roots is 
$$f'' \quad\text{change sign}\iff a>0$$
and the roots of $f''$ are $\alpha=-\sqrt{\frac{a}{3}}$ and $-\alpha$ and if we see the variations of $f'$ we find
$$f \quad \text{has 4 real roots}\iff a>0,\quad f'(\alpha)>0,\quad f'(-\alpha)<0$$
Edit
We have 
$$f'(\alpha)=8\left(\frac{a}{3}\right)^{3/2}+1>0\quad \forall a>0$$
and we can find easily (except miscalculation) that
$$f'(-\alpha)<0\iff a>\frac{3}{4}$$
so we conclude
$$f \quad \text{has 4 real roots}\iff a>\frac{3}{4} $$
A: First transform this equation into the equation with unknown $a$:
$a^{2}-(2x^{2} + 1)a + x^{4} + x = 0$
Solving the equation with unknown obtain two second degree equations:
$x^{2} + x - a = 0 $ and $ x^{2} - x + 1 - a = 0$
For $a < - \frac{1}{4}$ equation has no real roots
For $a = - \frac{1}{4}$ equation has two equal real roots
For $- \frac{1}{4} < a < \frac{3}{4}$ equation has two different real roots
For $a=\frac{3}{4}$ equation has real four roots of which three are equal
For $ a> \frac{3}{4}$ equation has four different real roots
