Task B-1.6. Depending on the real parameter $m$, determine the solution $x$ of the equation

$$
(m+x)^{2}-(x-3)^{2}=x\left(3+m^{2}\right)
$$
## Solution.

$$
\begin{aligned}
(m+x)^{2}-(x-3)^{2} & =x\left(3+m^{2}\right) \\
m^{2}+2 m x+x^{2}-x^{2}+6 x-9 & =3 x+m^{2} x \\
2 m x-m^{2} x+3 x & =9-m^{2} \\
x\left(3+2 m-m^{2}\right) & =9-m^{2} \\
x\left(3+3 m-m-m^{2}\right) & =(3-m)(3+m) \\
x(3(1+m)-m(1+m)) & =(3-m)(3+m) \\
x(1+m)(3-m) & =(3-m)(3+m)
\end{aligned}
$$

(2 points)

For $m \neq -1, m \neq 3$, the given equation has a unique solution

$$
x=\frac{(3-m)(3+m)}{(1+m)(3-m)}=\frac{3+m}{1+m}
$$

For $m=-1$, from $\left(^{*}\right)$ it follows that $0=8$, so the given equation has no solution.

(2 points)

For $m=3$, from $\left(*_{*}^{*}\right)$ it follows that $0=0$, so the given equation is indeterminate and every real number $x$ is a solution.

(2 points)