## Task B-3.2.

Determine all values of the real parameter $a$ such that the equation $4^{x}-(a+3) 2^{x}+4(a-$ $1)=0$ has exactly one real solution.
## Solution.

If we introduce the substitution $2^{x}=t$, we will get the quadratic equation

$$
t^{2}-(a+3) t+4 a-4=0
$$

This equation must have only one positive solution.

This can happen in two cases:

1. Equation (*) has only one solution, which must be positive, i.e., $D=0$ and $t>0$. It follows that

$$
\begin{array}{ll}
(a+3)^{2}-4(4 a-4)=0 \text { and } \frac{a+3}{2}>0 & 1 \text { point } \\
\Leftrightarrow a^{2}-10 a+25=0 \text { and } a>-3 & \\
\Leftrightarrow a=5 \text { and } t=4 . & 1 \text { point }
\end{array}
$$

2. Equation (*) has two solutions, but only one is positive.

1 point

It must hold that: $D>0$ and $t_{1} \cdot t_{2} \leqslant 0$.

From the first case, we see that $D>0$ for all $a \in \mathbb{R}, a \neq 5$.

According to Vieta's formulas, $t_{1} \cdot t_{2}=4 a-4$ so it follows that $4 a-4 \leqslant 0 \Rightarrow a \leqslant 1$. 1 point

The final solution to the problem is $a \in(-\infty, 1] \cup\{5\}$.