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 $x&lt;0$ for the first condition and $x&gt;0$ for the second condition. But, I don't know how do I proceed from here?

Any help would be appreciated.
So far, so good! Now: if $x&gt;0$, then $t=2^x&gt;1$; and if $x&lt;0$, then $t=2^x&lt;1$. 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&hellip; 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
$$\begin{split}
t^2+2at+a+1=0 &amp;\iff (y+1)^2+2a(y+1)+a+1=0\\
&amp;\iff y^2+2(a+1)y+(3a+2)=0.
\end{split}$$

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 $3a+2&lt;0$.
But also don't forget that $t=2^x$ can't be negative! In other words, $t&gt;0$, and therefore $y=t-1&gt;-1$. 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 $f(-1)&gt;0$, i.e. $(-1)^2+2(a+1)\cdot(-1)+(3a+2)&gt;0$, which simplifies to $a+1&gt;0$.
In order for distinct real roots, $a^2-a-1&gt;0$.

Since $x_1&gt;0,2^{x_1}=t_1=-a+\sqrt{a^2-a-1}&gt;1$. 

Similarly, $x_2&lt;0\therefore 0&lt;2^{x_2}=t_2=-a-\sqrt{a^2-a-1}&lt;1$.