6. For what values of the parameter a does the equation $\left(x^{2}-a\right)^{2}+2\left(x^{2}-a\right)+(x-a)+2=0$ have exactly one solution? Specify the solution for the found values of the parameter a. (20 points)
Solution.

Transform the equation $\quad\left(x^{2}-a\right)^{2}+2\left(x^{2}-a\right)+(x-a)+2=0$ to

the form $a^{2}-a\left(2 x^{2}+3\right)+x^{4}+2 x^{2}+x+2=0$.

This is a quadratic equation in terms of the parameter a. Let's find its discriminant $D=\left(2 x^{2}+3\right)^{2}-4\left(x^{4}+2 x^{2}+x+2\right)=(2 x-1)^{2} \Rightarrow a=\left[\begin{array}{l}x^{2}+x+1, \\ x^{2}-x+2\end{array}\right.$

Solve the two new quadratic equations in terms of the variable x: $x^{2}+x+1-a=0 \Rightarrow x_{1,2}=\frac{-1 \pm \sqrt{4 a-3}}{2}$ $x^{2}-x+2-a=0 \Rightarrow x_{3,4}=\frac{1 \pm \sqrt{4 a-7}}{2}$.

Note that the equations have coinciding solutions at $a=\frac{7}{4}$.

The maximum possible value of the parameter for which the equation has 2 roots is $a=7 / 4$, at which the first root is $x_{1}=-3 / 2$, and the other roots coincide $x_{2}=x_{3}=x_{4}=1 / 2$.

The only solution $x_{1}=-1 / 2$ is obtained at $a=3 / 4$.

Answer: $a=0.75, \quad x_{1}=-0.5$.