
Problem 1. For which values of the real parameter $a$ the equation $\lg \left(4 x^{2}-(8 a-1) x+5 a^{2}\right)+x^{2}+(1-2 a) x+2 a^{2}=\lg \left(x^{2}-2(a+1) x-a^{2}\right)$ has exactly one root?


Solution: Write the equation in the form:

$$
\begin{aligned}
& \lg \left(4 x^{2}-(8 a-1) x+5 a^{2}\right)+\frac{4 x^{2}-(8 a-1) x+5 a^{2}}{3}= \\
& \lg \left(x^{2}-2(a+1) x-a^{2}\right)+\frac{x^{2}-2(a+1) x-a^{2}}{3}
\end{aligned}
$$

Since $\lg t+\frac{t}{3}$ is an increasing function this equality is equivalent to

$$
\left\lvert\, \begin{aligned}
& 4 x^{2}-(8 a-1) x+5 a^{2}=x^{2}-2(a+1) x-a^{2} \\
& x^{2}-2(a+1) x-a^{2}>0
\end{aligned}\right.
$$

The equation in the above system is equivalent to $f(x)=0$ where $f(x)=x^{2}-(2 a-1) x+2 a^{2}$. Substitution $x^{2}=2 a x-x-2 a^{2}$ in the inequality gives $x<-a^{2}$. Therefore we have to find those $a$ for which the equation $f(x)=x^{2}-(2 a-1) x+2 a^{2}=0$ has exactly one root such that $x<-a^{2}$. There are three cases to be considered:

1. $f(x)=0$ has two roots and $-a^{2}$ is between the roots. In this case $f\left(-a^{2}\right)<0 \Rightarrow a^{2}(a+1)^{2}<0$, which is impossible.
2. $f(x)=0$ has two roots one of which equals $-a^{2}$ and the second one is less than $-a^{2}$. in this case $f\left(-a^{2}\right)=0$ and so $a=-1$
or $a=0$. When $a=-1$ we obtain $x_{1}=-2, x_{2}=-1$, which implies that $a=-1$ is a solution. If $a=0$ we get $x_{1}=-1, x_{2}=0$, which implies that $a=0$ is also a solution.
3. $f(x)=0$ has one root which is less than $-a^{2}$. Since $D=0$ we get that $-4 a^{2}-4 a+1=0$, and so $a_{1 / 2}=\frac{-1 \pm \sqrt{2}}{2}$. It is easy to check that for both values of $a$ the corresponding root is less than $-a^{2}$.

Therefore the solutions are $a=-1, a=0, a=\frac{-1 \pm \sqrt{2}}{2}$.
