Find $\;$ $a$ $\;$ such that one of the roots of the equation $\;$ $x^2 - \dfrac{15}{4} x + a = 0$ $\;$ is the square of the other.
Given quadratic equation: $\;\;\;$ $x^2 - \dfrac{15}{4} x + a = 0$ $\;\;\; \cdots \; (1)$
Given: $\;$ One root of equation $(1)$ is the square of the other.
Let the roots of $(1)$ be $\alpha, \; \alpha^2$.
Sum of roots of equation $(1)$ $= \alpha + \alpha^2 = \dfrac{- \left(-15/4\right)}{1} = \dfrac{15}{4}$
i.e. $\;$ $\alpha^2 + \alpha - \dfrac{15}{4} = 0$
i.e. $\;$ $4 \alpha^2 + 4 \alpha - 15 = 0$
i.e. $\;$ $\left(2 \alpha + 5\right) \left(2 \alpha - 3\right) = 0$
i.e. $\;$ $\alpha = \dfrac{-5}{2}$, $\;\;$ or $\;\;$ $\alpha = \dfrac{3}{2}$
Product of roots of equation $(1)$ $= \alpha \times \alpha^2 = \alpha^3 = \dfrac{a}{1} = a$ $\;\;\; \cdots \; (2)$
Substituting $\alpha = \dfrac{-5}{2}$ in equation $(2)$ gives
$a = \left(\dfrac{-5}{2}\right)^3 = \dfrac{-125}{8}$
Substituting $\alpha = \dfrac{3}{2}$ in equation $(2)$ gives
$a = \left(\dfrac{3}{2}\right)^3 = \dfrac{27}{8}$