Finding all a to have atleast one real root 
Find all $a \in \Bbb R$ for which $$(1 + a) \left(\frac{x^2}{1 + x^2}\right)^2 - 3a\left(\frac{x^2}{1 + x^2}\right) + 4a = 0$$ has at least one real root.

For this problem, I tried substituting $y = x^2 / 1 + x^2$ and then arrived at the quadratic $$(1+a)y^2 - 3ay + 4a = 0$$ then the discriminant $$\Delta = 9a^2 - 4(1+a)(4a) = 9a^2 - 16a - 16 a^2 = -7a^2 - 16a$$ Now setting the discriminant $\geq 0$ since we need real roots gives $$-\frac{16}{7}\le\:a\le\:0$$.
However, I am confident this is an incorrect way since this considers the quadratic and not the original equation (??). I do not have access to the solution/answers so I am a bit lost on this.
So what is the correct approach for this? Any hints will suffice.
 A: Note that

*

*$\frac{x^2}{1 + x^2}$ takes values in $[0, 1)$.


*$\frac{x^2}{1 + x^2}$ takes all values in $[0, 1)$.
Thus, your original equation (in $x$) has a real solution iff your transformed equation (in $y$) has a solution in $[0, 1)$. (Note that the equation in $y$ may have two real solutions with only one being in $[0, 1)$, that is OK.)
What you have done will only ensure that the transformed equation has root(s) in $\Bbb R$. That is not enough.
Firstly note that if $a = -1$, then you have a linear equation and you can note that case simply.
Now, assume that $a \neq -1$. You can explicitly solve the quadratic in $y$ and impose the condition that one of the roots lies in $[0, 1)$. (As a first step, you must have that both roots are real. You have already shown that this restricts $a$ to be in $[-16/7, 0]$.)
A: Why to make the problem more complex than it is ?
Since, in the real domain, $(1+x^2) >0$, multiply everything by $(1+x^2)^2$ to end with
$$(1+2a)x^4+5ax^2+4a=0$$ which is a quadratic in $x^2$. So, you know the condition for real roots in $x^2$ and at leat one of these roots must be non-negative.
