Problem 10.7. Nikita schematically drew the graph of the quadratic polynomial $y=a x^{2}+b x+c$. It turned out that $A B=C D=1$. Consider four numbers $-a, b, c$ and the discriminant of the quadratic polynomial. It is known that three of them are equal in some order to $1 / 4, -1, -3 / 2$. Find what the fourth number is.

![](https://cdn.mathpix.com/cropped/2024_05_06_62ad5020cf8c9b2474f9g-4.jpg?height=258&width=418&top_left_y=1500&top_left_x=518)
Answer: $-1 / 2$.

Solution. From the drawn graph, it follows that $a$, $b$, and $c$ are negative. Indeed, $a$ is negative because the branches of the parabola are directed downwards. The free term $c$ is negative because it equals the value of the trinomial at $x=0$. Finally, $b$ coincides with the slope of the line $b x+c$, which is tangent to the graph of the trinomial at the point with abscissa $x=0$; it is not difficult to see that this coefficient is also negative. (Alternatively, one can notice that the abscissa of the vertex of the parabola is negative, and it equals $-\frac{b}{2 a}$, which determines the sign of $b$.

Since among the three numbers given to us, one is positive, it can only be the discriminant $d=1 / 4$. Consider the distance $A B$ between the roots of the trinomial. By the condition, it equals 1. On the other hand, subtracting the formulas for the roots of the quadratic trinomial from each other, one can express this distance in terms of $a$ and $d$:

$$
1=A B=\frac{\sqrt{d}}{|a|}=\frac{1}{2|a|}
$$

We get $|a|=1 / 2$, that is, $a=-1 / 2$. This number was not in the list, so this is the answer.