Problem 9.2. Find all values of $a$, for which the equation

$$
\frac{2 a}{(x+1)^{2}}+\frac{a+1}{x+1}-\frac{2(a+1) x-(a+3)}{2 x^{2}-x-1}=0
$$

has two real roots $x_{1}$ and $x_{2}$ satisfying the relation $x_{2}^{2}-a x_{1}=a^{2}-a-1$.

Ivan Landjev
9.2. The given equation is equivalent to

$$
a x^{2}+(1-2 a) x+(1-a)=0
$$

where $x \neq-1,-\frac{1}{2}, 1$. Hence this equation has two real roots $x_{1}$ and $x_{2}$ such that

$$
x_{2}^{2}-a x_{1}=a^{2}-a-1
$$

Since $x_{1}+x_{2}=\frac{2 a-1}{a}$ we get that

$$
x_{2}^{2}+a x_{2}-a^{2}-a+2=0
$$

This together with the identity

$$
a x_{2}^{2}+(1-2 a) x_{2}+1-a=0
$$

implies that

$$
\left(a^{2}+2 a-1\right) x_{2}=a^{3}+a^{2}-3 a+1=\left(a^{2}+2 a-1\right)(a-1)
$$

The coefficient of $x_{2}$ vanishes if $a=-1 \pm \sqrt{2}$.

If $a=-1+\sqrt{2}$, then

$$
(-1+\sqrt{2}) x_{2}^{2}+(3-2 \sqrt{2}) x_{2}+(2-\sqrt{2})=0
$$

which is impossible, since the discriminant of this quadratic equation equals $33-24 \sqrt{2}<0$, i.e. it has no real roots.

If $a=-1-\sqrt{2}$ we get the equation

$$
(-1-\sqrt{2}) x_{2}^{2}+(3+2 \sqrt{2}) x_{2}+(2+\sqrt{2})=0
$$

that has two real roots, which are not equal to $\pm 1$ and $-\frac{1}{2}$.

Let now $a \neq-1 \pm \sqrt{2}$. Then $x_{2}=a-1$ and hence $a(a-1)(a-3)=0$. Since $a \neq 0,1$ we get $a=3$. In this case the roots of the given equation are $-\frac{1}{3}$ and 2 , and they satisfy the given condition.

Thus the desired values of $a$ are $-1-\sqrt{2}$ and 3 .