Find $a\in \mathbb{R}$ for which $a\cdot \left(\frac{1}{1+x^2}\right)^2-3\cdot\frac{a}{1+x^2}+1=0$ will have all roots imaginary 
Find $a\in  \mathbb{R}$ for which $a\cdot \left(\frac{1}{1+x^2}\right)^2-3\cdot\frac{a}{1+x^2}+1=0$ will have all roots imaginary.

My attempt is as follows:-
Let $t=\frac{1}{1+x^2}$, and let's find out its range for which x is imaginary
$$t=\frac{1}{1+x^2}$$
$$(1+x^2)\cdot t=1$$
$$tx^2+t-1=0$$
$$D<0$$
$$0-4t(t-1)<0$$
$$t(t-1)>0$$
$$t\in (-\infty,0)\quad \cup \quad (1,\infty)$$
So for the equation $at^2-3at+1=0$, we have to find such values of a for which $t\in (-\infty,0) \cup  (1,\infty)$. As $t$ should be real,so
$$D\geq 0\Leftrightarrow 9a^2-4a\geq 0\Leftrightarrow a(9a-4)\geq0$$
$$a\in \left(-\infty,0\right] \cup \left[\frac{4}{9},\infty\right)$$
But if we place $a=0$ in the quadratic equation in $t$, then $0+0+1=0$, which is not possible hence $a\in \left(-\infty,0\right) \cup \left[\frac{4}{9},\infty\right)$.
Now as we know that roots of quadratic equation $at^2-3at+1=0$ should lie in $(-\infty,0)\cup  (1,\infty)$. So
Case 1 : When both roots are negative
$$af(0)>0$$
$$a>0$$
$0$ is greater than both the roots, so 
$0>(a+b)/2$ where a and b are roots.
$$0>\frac{3a}{2a}$$
$$0>\frac{3}{2}$$
So $a\in \phi$ for first case
Case 2: When both roots are greater than $1$
$$af(1)>0$$
$$a(a-3a+1)>0$$
$$a(2a-1)<0$$
$$a\in \left(0,\frac{1}{2}\right)$$
$1$ should lie before the roots on the x-axis, so $1<\frac{a+b}{2}$
$$1<\frac{3a}{2a}$$
$$1<\frac{3}{2}$$
So $a\in \left(0,\frac{1}{2}\right)$ for the second case
Case 3: When one root is greater than $1$ and another is negative:
$$af(0)<0\quad \cap \quad af(1)<0$$
$$a<0\quad \cap\quad a(a-3a+1)<0$$
$$a<0\quad \cap \quad a(2a-1)>0$$
$$a\in \left(-\infty,0\right)$$
Hence $a\in \left(-\infty,0\right) \cup \left[\frac{4}{9},\frac{1}{2}\right)$ 
but answer is $a\in \left(-\infty,\frac{1}{2}\right)$
What mistake I am doing, I thought about it a lot but didn't get any breakthroughs. Please help me in this.
 A: In your proof you have excluded the interval $(0,4/9)$. Why? For example if $a=1/3\in(0,4/9)$ then the equation becomes
$$\frac{1+3x^2+3x^4}{(1+x^2)^2}=0$$
which has not real roots because the l.h.s. is always positive. Hence $1/3$ should be included in the required set.
If $z(x)=1/(1+x^2)$ then $z(\mathbb{R})=(0,1]$. Let $p(z)=az^2-3az+1$ then
we have to find for which real $a$, $p((0,1])$ does not contain $0$.
We have $3$ cases according to the sign of $a$.
1) If $a>0$ then $p$ is decreasing in $(-\infty,3/2)$ and $p((0,1])=[-2a+1,1)$ and therefore $0\not \in [-2a+1,1)$ iff $-2a+1>0$ iff $a<1/2$.
2) If $a<0$ then $p$ is increasing in $(-\infty,3/2)$ and $p((0,1])=(1,-2a+1]$ and therefore $0\not \in (1,-2a+1]$ for all  $a<0$.
3) If $a=0$ then $p$ is identically $1$ and therefore it is never zero.
We may conclude that the given equation has not real roots if and only if $a\in \left(-\infty,\frac{1}{2}\right)$.
A: Hint: Factorizing your equation we get
$$x^4+x^2(2-3a)+1-2a=0$$
You can also write $$at^2-3at+1=0$$ where $t=\frac{1}{1+x^2}>0$
$a=0$ is impossible, so we get by the quadratic formula
$$t_{1,2}=-\frac{3}{2}\pm\sqrt{\frac{9}{4}-\frac{1}{a}}$$
If $$\frac{9}{4}-\frac{1}{a}<0$$ then the roots are imaginary.
