
Problem 10b.1. Find all values of the real parameter $a$ such that the equation

$$
\log _{x}\left(x^{2}+x+a\right)^{2}=4
$$

has unique solution.


Solution: The equation is equivalent to $\left(x^{2}+x+a\right)^{2}=x^{4}, x>$ $0, x \neq 1$. Further, $\left(x^{2}+x+a\right)^{2}=x^{4} \Longleftrightarrow(x+a)\left(2 x^{2}+x+\right.$ $a)=0$ with roots $x_{1}=-a$ and $x_{2 / 3}=\frac{-1 \pm \sqrt{1-8 a}}{4}$, provided $1-8 a \geq 0$. Since $\frac{-1-\sqrt{1-8 a}}{4}0$ and $x_{2}>0$. Note that $x_{1}=x_{2}$ implies $a=0$ which is a contradiction. Therefore if the equation has unique solution then it is necessary to have $a<0$ and one of the two roots equals 1 . Thus, $x_{1}=1 \Rightarrow a=-1$ and $x_{2}=1 \Rightarrow a=-3$.
