\section*{Problem 1 - 061021}

Determine all real numbers \(a\) for which one of the roots of the quadratic equation

\[
x^{2}-\frac{15}{4} x+a=0
\]

is the square of the other root!
}

Let the two roots of the quadratic equation be \(w\) and \(w^{2}\). Then, according to Vieta's formulas:

\[
w^{2}+w=\frac{15}{4} \quad(1) \quad ; \quad w^{2} \cdot w=a
\]

From (1) it follows that

\[
w_{1}=\frac{3}{2} \quad \text { and } \quad w_{2}=-\frac{5}{2}
\]

Due to (2), we have

\[
a_{1}=w_{1}^{3}=\frac{27}{8} \quad \text { and } \quad a_{2}=w_{2}^{3}=-\frac{125}{8}
\]

By substitution, it can be found that the determined values indeed satisfy the conditions:

\[
x^{2}-\frac{15}{4} x+\frac{27}{8}=0
\]

has the roots \(x_{1}=\frac{9}{4}\) and \(x_{2}=\frac{3}{2}\).

\[
x^{2}-\frac{15}{4} x-\frac{125}{8}=0
\]

has the roots \(x_{1}=\frac{25}{4}\) and \(x_{2}=-\frac{5}{2}\). The given condition is satisfied by \(a_{1}=\frac{27}{8}\) and \(a_{2}=-\frac{125}{8}\) and only by these.