Condition on $a$ for $(x^2+x)^2+a(x^2+x)+4=0$ Find the set of values of $a$ if $$(x^2+x)^2+a(x^2+x)+4=0$$ has
$(i)$ All four real and distinct roots
$(ii)$ Four roots in which only two roots are real and distinct.
$(iii)$ All four imaginary roots
$(iv)$ Four real roots in which only two are equal.
Now if I set $x^2+x=t$ then even if $t^2+at+4=0$ has real roots in is not necessary that $(x^2+x)^2+a(x^2+x)+4=0$ will have real roots too.  So how to derive the condition on a?
Could someone give me some direction?
 A: Suppose we want to find the values of $a$ such that there are four real and distinct roots (values of $x$ such that the equation is true). 
This means that if we let $x^2+x=t$ as you did, then we want the two possible values of $t$ to be distinct. Call them $t_1$ and $t_2$. We want that the following two equations both have two roots, and that all of the roots are distinct:
$$x^2+x-t_1=0$$
$$x^2+x-t_2=0$$
The quadratic formula gives
$$x=\frac{-1\pm\sqrt{1+4t_1}}{2}$$
$$x=\frac{-1\pm\sqrt{1+4t_2}}{2}$$
This means we want $t_1\neq t_2$, and $$t_1,t_2 > -\frac{1}{4}$$
Note that this inequality is strict, since a determinant of $0$ would give a double root. 

We can now return to the equation involving $a$. We want to find which $a$ allow the two roots of the following equation to satisfy $t_1\neq t_2$ and $t_1,t_2 > -1/4$.
$$t^2+at+4=0$$
We use the quadratic formula again:
$$t_1=\frac{-a + \sqrt{a^2-16}}{2}$$
$$t_2=\frac{-a - \sqrt{a^2-16}}{2}$$
Since we want two distinct real values of $t$, $a$ must have absolute value strictly greater than $4$, to ensure a strictly positive determinant.
We also want $a$ such that both of these inequalities hold:
$$\frac{-a + \sqrt{a^2-16}}{2} > -\frac{1}{4}$$
$$\frac{-a - \sqrt{a^2-16}}{2} > -\frac{1}{4}$$
The second inequality implies the first, so it's sufficient that the second holds. It's equivalent to
$$2a < 1 - 2\sqrt{a^2-16}$$
Clearly all values of $a$ greater than $4$ don't work. We're left with $a < -4$. 
At $a=-4$, $$1-2\sqrt{a^2}-16 > 2a$$
Let $f(a)=1-2\sqrt{a^2-16}$ and $g(a)=2a$. Notice that $f(a)$ is greater than $g(a)$ at $a=-4$. The equation $f(a)=g(a)$ has no solutions (to see why, subtract one from both sides and square).  This means $f(a)$ never crosses $g(a)$, and since both functions are continuous on $a \in (-\infty, 4]$, we know that $f(a)$ is greater than $g(a)$ for all $a\leq -4$. 
The values of $a$ such that there are four real and distinct roots are all $a$ such that $\boxed{a < -4}$.

This same method (using the quadratic formula first on $x^2+x=t$, then on $t^2+at+4=0$, doing casework at each step) can be applied to the other parts of your question. Good luck!
A: Alternatively, graphing it:
Let $a=y$, then $(x^2+x)^2+y(x^2+x)+4=0$
$\displaystyle y=\frac{1}{4}-\left( x+\frac{1}{2} \right)^{2}+
\frac{4}{\frac{1}{4}-\left( x+\frac{1}{2} \right)^{2}}$

