[ Quadratic equations. Formula for roots ] [ Quadratic equations. Vieta's theorem ]

For what values of the parameter $a$ is one of the roots of the equation $x^{2}-15 / 4 x+a^{3}=0$ the square of the other?
According to Vieta's formula, the free term is equal to the cube of the specified root. Therefore, the roots of the equation are $a$ and $a^{2}$.

Thus, $a+a^{2}=15 / 4$, from which

$a=\frac{3}{2}$ or $a=-\frac{5}{2}$.

## Answer

When $a=1.5$ and $a=-2.5$.