## Problem II - 4

Calculate the numbers $p$ and $q$ such that the roots of the equation

$$
x^{2} + p x + q = 0
$$

are $D$ and 1 - $D$, where $D$ is the discriminant of this quadratic equation.
## Solution

From the formulas that relate the roots and the coefficients of the equation, we have

$$
\begin{aligned}
& D+1-D=-p \\
& D \cdot(1-D)=q
\end{aligned}
$$

From the first, we immediately get $p=-1$; from the second, $q=D-D^{2}$.

But $D=p^{2}-4 q=1-4 q=1-4\left(D-D^{2}\right)$, that is,

$$
4 D^{2}-5 D+1=0
$$

so $D=1$ or $D=1 / 4$.

If $D=1$, then $q=1-1^{2}=0$.

If $D=1 / 4$, then $q=\frac{1}{4}-\left(\frac{1}{4}\right)^{2}=\frac{3}{16}$.

The obtained values are

$$
\begin{aligned}
& (p, q)=(-1,0) \\
& (p, q)=\left(-1, \frac{3}{16}\right)
\end{aligned}
$$

Finally, it is verified by direct substitution that in these two cases the equation has roots $D$ and $-D$.