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 in the trinomial is increased by 1, the resulting quadratic trinomial will have exactly one root.
Answer: $\frac{1}{8} \cdot x^{2}-\frac{3}{4} \cdot x+\frac{1}{8}$.

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

$$
b^{2}-4(a+1) c=(b+1)^{2}-4 a c=b^{2}-4 a(c+1)=0
$$

From the fact that the first and third expressions are zero, it follows that $a=c$. Then from the equality of the second expression to zero, we get that $b+1= \pm 2 a$. After eliminating $b$ and from the last equality, we get the equation $0=(-1 \pm 2 a)^{2}-4 a(a+1)=1 \mp 4 a-4 a$. Therefore, the sign 干 is a minus. Then $c=a=\frac{1}{8}$ and $b=-1+2 \cdot \frac{1}{8}=-\frac{3}{4}$.