Problem 10.4. Given a quadratic trinomial $P(x)$. It is known that the equations $P(x)=x-2$ and $P(x)=1-x / 2$ have exactly one root each. What is the discriminant of $P(x) ?$
Answer: $-1 / 2$.

Solution. Let $P(x)=a x^{2}+b x+c, a \neq 0$. Consider the first quadratic equation and its discriminant:

$$
\begin{gathered}
a x^{2}+b x+c=x-2 \\
a x^{2}+x(b-1)+(c+2)=0 \\
D_{1}=(b-1)^{2}-4 a(c+2)
\end{gathered}
$$

Now consider the second quadratic equation and its discriminant:

$$
a x^{2}+b x+c=1-x / 2
$$

$$
\begin{gathered}
a x^{2}+x(b+1 / 2)+(c-1)=0 \\
D_{2}=(b+1 / 2)^{2}-4 a(c-1)
\end{gathered}
$$

Since the quadratic equations have one root each, both of these discriminants are equal to zero. We obtain the system

$$
\left\{\begin{array}{l}
(b-1)^{2}-4 a(c+2)=0 \\
(b+1 / 2)^{2}-4 a(c-1)=0
\end{array}\right.
$$

Expanding the brackets, we get

$$
\left\{\begin{array}{l}
b^{2}-2 b+1-4 a c-8 a=0 \\
b^{2}+b+1 / 4-4 a c+4 a=0
\end{array}\right.
$$

Adding the first equation and twice the second, we get

$$
\begin{gathered}
b^{2}-2 b+1-4 a c-8 a+2 \cdot\left(b^{2}+b+1 / 4-4 a c+4 a\right)=0 \\
3 b^{2}-12 a c+3 / 2=0 \\
b^{2}-4 a c=-1 / 2
\end{gathered}
$$

And this is the discriminant of $P(x)$.