Find the exhaustive values of "a", given the condition 
If the equation $2^{2x} + a*2^{x+1} + a + 1=0$ has roots of opposite sign then the exhaustive values of a are?

I tried taking $2^x = t$. But then didn't know what to do.
The equation became, $t^2 + 2at + a +1 =0$. But then, what conditions should I impose?
Since one root is negative and the other positive, the only conclusion that I could draw was that $x<0$ for the first condition and $x>0$ for the second condition. But, I don't know how do I proceed from here?
Any help would be appreciated.
 A: So far, so good! Now: if $x>0$, then $t=2^x>1$; and if $x<0$, then $t=2^x<1$. So for your new equation $t^2+2at+a+1=0$ you want it to have two roots $t$ of which one is greater and one is less than $1$. Comparing roots with $1$ seems a bit difficult; comparing with $0$ would be easier… So let's make another change of variables: $t$ is greater than or less than $1$ if and only if $y=t-1$ is greater than or less than $0$ respectively. Substituting $y=t-1$, i.e. $t=y+1$, the quadratic equations transforms into
$$\begin{split}
t^2+2at+a+1=0 &\iff (y+1)^2+2a(y+1)+a+1=0\\
&\iff y^2+2(a+1)y+(3a+2)=0.
\end{split}$$
Now, we have to satisfy two conditions:


*

*A quadratic equation (with real coefficients and the leading term of $1$, as is the case here) has roots of opposite signs if and only if its constant term is negative, so we need $3a+2<0$.

*But also don't forget that $t=2^x$ can't be negative! In other words, $t>0$, and therefore $y=t-1>-1$. For a parabola $f(y)=y^2+2(a+1)y+(3a+2)$ opening up with one positive and one negative root, we need to guarantee that even the negative root is greater than $-1$, or in other words, that $-1$ lies to the left of both roots. This conditions will be satisfied if $f(-1)>0$, i.e. $(-1)^2+2(a+1)\cdot(-1)+(3a+2)>0$, which simplifies to $a+1>0$.

A: The hint:
Solve the following system.
$$1^2+2a\cdot1+a+1<0$$ and
$$a+1>0.$$
The first inequality says that $1$ is placed between $2^{x_1}$ and $2^{x_2}.$
The second inequality says that the smaller root of the quadratic equation is positive.
A: In order for distinct real roots, $a^2-a-1>0$.
Since $x_1>0,2^{x_1}=t_1=-a+\sqrt{a^2-a-1}>1$. 
Similarly, $x_2<0\therefore 0<2^{x_2}=t_2=-a-\sqrt{a^2-a-1}<1$.
