$\mathrm{Az}$

$$
\frac{x^{2}+p}{x}=-\frac{1}{4}
$$

equation, whose roots are $x_{1}$ and $x_{2}$, determine $p$ such that
a) $\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{1}}=-\frac{9}{4}$,

b) one root is 1 less than the square of the other root.
Our equation can also be written as

$$
4 x^{2}+x+4 p=0
$$

a) It is known that for the quadratic equation $a x^{2}+b x+c=0$, the elementary symmetric forms of the roots are $x_{1}+x_{2}=-\frac{b}{a}$ and $x_{1} x_{2}=\frac{c}{a}$.

$$
\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{1}}=\frac{x_{1}^{2}+x_{2}^{2}}{x_{1} x_{2}}=\frac{\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}}{x_{1} x_{2}}=\frac{\frac{b^{2}}{a^{2}}-\frac{2 c}{a}}{\frac{c}{a}}=\frac{b^{2}-2 a c}{a c}
$$

In this case, $a=4, b=1$ and $c=4 p$, so

$$
\frac{x_{1}}{x_{2}}+\frac{x^{2}}{x_{1}}=\frac{1-32 p}{16 p}=-\frac{9}{4}
$$

from which

$$
p=-\frac{1}{4}
$$

b) According to the problem, the roots are $x_{1}$ and $x_{1}^{2}-1$.

# Therefore

$$
x_{1}+\left(x_{1}^{2}-1\right)=-\frac{b}{a}=-\frac{1}{4}
$$

and

$$
x_{1}\left(x_{1}^{2}-1\right)=\frac{c}{a}=p
$$

(1) is a quadratic equation in $x_{1}$, from which $x_{1}$ can be calculated. For simplicity, let's denote $x_{1}$ by $y$, then

$$
y^{2}+y-1=-\frac{1}{4}
$$

which is

$$
4 y^{2}+4 y-3=0
$$

from which

$$
y_{1}=\frac{1}{2}, \quad y_{2}=-\frac{3}{2}
$$

Substituting these values of $y=x_{1}$ into (2),

$$
\begin{gathered}
p_{1}=\frac{1}{2}\left(\frac{1}{4}-1\right)=\frac{1}{2}\left(-\frac{3}{4}\right)=-\frac{3}{8} \\
p_{2}=\left(-\frac{3}{2}\right)\left(\frac{9}{4}-1\right)=\left(-\frac{3}{2}\right) \frac{5}{4}=-\frac{15}{8} .
\end{gathered}
$$

Pál Bártfai (Bp., I., Petőfi g. II. o. t.)