[ Methods for solving problems with parameters ] [ Quadratic equations. Vieta's theorem ]

For what values of the parameter $a$ does the equation $(a-1) x^{2}-2(a+1) x+2(a+1)=0$ have only one non-negative solution

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When $a=1$, the equation becomes linear: $4-4 x=0$, the only root of which is $x=1$.

Let $a \neq 1$. The equation can have one repeated non-negative root or two distinct roots, one of which is non-negative.

In the first case, the discriminant $(a+1)^{2}-2(a-1)(a+1)=-a^{2}+2 a+3=0$, that is, $a=-1$ or $a=3$. When $a=-1$, the equation becomes $2 x^{2}=0$ and has a single root $x=0$. When $a=3$, the equation becomes $2(x-2)^{2}=0$ and has a single root $x=2$.

In the second case, the discriminant $-a^{2}+2 a+3>0 \quad \Leftrightarrow-1<a<3, \quad a \neq 1$.

Under this condition, the product of the roots cannot be zero (since the constant term is not zero), so exactly one of the roots is non-negative if and only if the product of the roots is negative:

$\frac{2(a+1)}{a-1}<0 \Leftrightarrow-1<a<1$

## Answer

$-1 \leq a \leq 1, a=3$.