Find the value of a for which the equation $x^4 - ax^2 + 9 = 0$ has four real and distinct roots.  Q. Find the value of a for which the equation $x^4 - ax^2 + 9 = 0$ has four real and distinct roots.
I have attempted this question in the following way:
Let $x^2 = t$.
So the given equation becomes $t^2 - at + 9 = 0$ --> (Equation 1)
Since we need the roots to be real and distinct:
$D > 0$
So, $a^2 - 36 > 0$ (from Equation 1)
That implies, $a\,\epsilon\, (-\infty, -6) \cup (6,\infty)$. 
But the answer given is $a\,\epsilon\, (6,\infty)$. So where am I wrong?
Also if you can help me with the intervals in which a lies for no real roots and only two real roots it would be helpful.
 A: HINT: the condition $a^2-36>0$ implies that there are two real and distinct roots $t $ for equation 1, but the question talks about four real and distinct roots $x$ for the initial equation.
A: Here's a different approach. If the polynomial has four real roots, then its derivative has to have three roots. The derivative is 
$$
4x^3-2ax=2x(2x^2-a).
$$
This will have three roots ($0$, $\pm\sqrt{a/2}$) as long as $a>0$. 
Now, the polynomial will have four roots if we can push the minimum below the $x$-axis (note that the polynomial is symmetric with respect to the $y$-axis). The minima occur at $\pm\sqrt{a/2}$ (because the second derivative is $12x^2-2a$, so it is positive at both). The value of the polynomial at $\pm \sqrt{a/2}$ is 
$$
\frac{a^2}{4}-\frac{a^2}2+9=9-\frac{a^2}4.
$$
For this to be negative we need $a^2>36$, so $a>6$. 
A: If $a < 0$, Descartes' Rule of Signs tells you $x^4 - a x^2 + 9$ has no positive real roots and no negative real roots.  The only way out is if it has four roots at zero (it doesn't), but these wouldn't be distinct.
