2. Find all quadratic trinomials $a x^{2}+b x+c$ with real coefficients $a, b$, and $c$, such that if any one of the three coefficients is replaced by 1, the resulting quadratic trinomial will have exactly one root.
Answer: $\frac{1}{2} \cdot x^{2} \pm \sqrt{2} \cdot x+\frac{1}{2}$.

Solution. According to the condition, the quadratic trinomials $x^{2}+b x+c, a x^{2}+x+c$ and $a x^{2}+b x+1$ each have exactly one root. Therefore, their discriminants are zero. Thus,

$$
b^{2}-4 c=1-4 a c=b^{2}-4 a=0 .
$$

From the fact that the first and third expressions are equal to zero, it follows that $a=c>0$. Then, from the equality of the second expression to zero, we get that $a=c=\frac{1}{2}$. Therefore, $b= \pm \sqrt{2}$.