Bulgarian Mathematical Competitions

Solution:
If $a x+2<0$, then the equation has no real roots. If $a x+2 \geq 0$, it is equivalent to $a x^{2}+a x+2=(a x+2)^{2}$, i.e., $(a^{2}-a) x^{2}+3 a x+2=0$. The last equation has a unique real root in the following three cases.

Case 1. The coefficient of $x^{2}$ vanishes and the respective linear equation has a root $x$ such that $a x+2 \geq 0$.
If $a=0$, then $2=0$ which is impossible. If $a=1$, then $x=-\frac{2}{3}$ and $a x+2=-\frac{2}{3}+2=\frac{4}{3}>0$. Hence $a=1$ is a solution of the problem.

Case 2. The coefficient of $x^{2}$ is non-zero, i.e., $a \neq 0,1$, and the respective quadratic equation has a unique real root $x$ with $a x+2 \geq 0$. Then $D=9 a^{2}-8(a^{2}-a)=a^{2}+8 a=0$ and hence $a=-8$. Then $x=\frac{1}{6}$ and $a x+2= -8 \cdot \frac{1}{6}+2=\frac{2}{3}>0$, i.e., $a=-8$ is a solution of the problem.

Case 3. The coefficient of $x^{2}$ is non-zero, i.e., $a \neq 0,1$, and the respective quadratic equation has two real roots $x_{1}<x_{2}$ such that $a x_{1}+2<0 \leq a x_{2}+2$, i.e., $-\frac{2}{a} \in\left(x_{1}, x_{2}\right]$.
If $-\frac{2}{a}=x_{2}$, then $(a^{2}-a)\left(-\frac{2}{a}\right)^{2}+3 a\left(-\frac{2}{a}\right)+2=0$. Hence $-\frac{1}{a}=0$, contradiction. Therefore $-\frac{2}{a} \in\left(x_{1}, x_{2}\right)$ which is equivalent to
$$(a^{2}-a)\left((a^{2}-a)\left(-\frac{2}{a}\right)^{2}+3 a\left(-\frac{2}{a}\right)+2\right)<0$$
It is easy to see that the solutions of the above inequality are $a>1$. So, the given equation has a unique real root for $a=-8$ and $a \geq 1$.