<s>[INST] Solve the equation $\sqrt{a(2^{x}-2)+1}=1-2^{x}$ for every value of the parameter $a$. Question : Solve the equation $\sqrt{a(2^{x}-2)+1}=1-2^{x}$   for every value of the parameter a.
I have solved the problem as follows
$\sqrt{a\left(2^{x}-2\right)+1}=1-2^{x}$
$a\left(2^{x}-2\right)+1=\left(1-2^{x}\right)^{2}=2^{2 x}+1-2 \cdot 2^{x}=2^{2 x}-2^{x+1}+1$
$a 2^{x}-2 a=2^{2 x}-2^{x+1}$
$2^{2 x}-(a-1) 2^{x}+2 a=0$
$y^{2}-(a-1) y+2 a=0$
$y=\frac{(a-1) \pm \sqrt{(a-1)^{2}-8 a}}{2}=\frac{(a-1) \pm \sqrt{a^{2}-10 a+1}}{2}$
$2^{x}=\frac{(a-1) \pm \sqrt{(a-5)^{2}-24}}{2}$
After this there are so many conditions on a. Do i need to check for each and every value ?
 [/INST] By letting $y=1-2^x$, you can simplify the problem to:
$$
\sqrt{a(3-y)+1} = y
$$
Since root is non-negative, $y\ge0$. Since $2^x$ is always positive, $y<1$. Thus we want to find the solution of the equation above with the condition $0\le y<1$.
Since both sides are non-negative, we can square them and keep the equality:
$$
a(3-y)+1=y^2\\
y^2+ay - (1+3a)=0
$$
This equation will have the solution when $D=a^2+4(1+3a)\ge 0$.
By solving the corresponding equation, we will get $a\in I_1=(-\infty;-6-4\sqrt2]\cup[-6+4\sqrt2;+\infty)$.
Solution for $y$:
$$
y_{1,2}=\frac12\left(-a\pm\sqrt{a^2+12a+4}\right).
$$
Now let's check when $y\ge0$. For smaller root:
$$
y_1\ge0\\
-a-\sqrt{a^2+12a+4} \ge 0\\
-a \ge \sqrt{a^2+12a+4}\\
a\le0\qquad\text{and}\qquad a^2\ge a^2+12a+4
$$
So, smaller root is always smaller than 0. For larger root:
$$
y_2\ge 0\\
-a+\sqrt{a^2+12a+4} \ge 0 \\
\sqrt{a^2+12a+4}\ge a\\
\text{either}\qquad a<0\qquad \text{or }\qquad a^2+12a+4\ge a^2 \\
\text{either}\qquad a<0\qquad \text{or }\qquad a\ge -1/3
$$
Thus, $y_2\ge 0$ when exists.
Finally, we need to check that $y<1$. We don't need to check the $y_1$, since it already failed previous condition. For larger root:
$$
y_1 < 1\\
-a+\sqrt{a^2+12a+4} < 2\\
\sqrt{a^2+12a+4} < 2+a\\
a\ge-2\qquad\text{and}\qquad a^2+12a+4 < (2+a)^2 \\
a\ge-2\qquad\text{and}\qquad 12a+4 < 4a+4 \\
a\ge-2\qquad\text{and}\qquad a< 0 \\
$$
Thus, $a\in I_1 \cap[-2; 0)=[-6+4\sqrt2; 0)$. Finally, the answer:
When $a\in [-6+4\sqrt2; 0)$, the equation has one root, which is:
$$
x = \log_2\left(1-\frac{\sqrt{a^2+12a+4}-a}2\right).
$$
Note. I didn't include analysis of that expression under the root should be non-negative when was calculating $y\ge0$ and $y<1$, because we already did this analysis, and the corresponding interval would be intersecting the result anyway.
Note 2. We didn't need to check that $a(3-y)+1\ge 0$, because we set it equal to $y^2$ which is always non-negative
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