If one of the roots of the equation $x^2+p x+q=0$ is equal to the square of the other then
If one of the roots of the equation $x^2+p x+q=0$ is equal to the square of the other then
$p\left(q^2-3 p\right)=q(p-1)$
$p\left(3 p-q^2\right)=p(p+1)$
$p\left(3 q-p^2\right)=q(q-1)$
$p\left(3 q-p^2\right)=q(q+1)$
Solution
Given equation $x^2+p x+q=0$ ...(i)
On comparing with $a x^2+b x+c=0$
$a=1, b=p, c=q$
Let $\alpha, \beta$ are the roots of Eq. (i) and given that $\alpha=\beta^2$
$\therefore \quad \alpha+\beta=\frac{-b}{a}=-p$ and $\alpha \beta=\frac{c}{a}=q$
$\Rightarrow \quad \beta^2+\beta=-p$ and $\beta^2 \cdot \beta=q$
and $\quad\beta^3=q$ ...(ii)
and $\quad \beta=(q)^{1 / 3}$
On taking cube both sides
$\left(\beta^2+\beta\right)^3=(-p)^3$
$\Rightarrow \quad \beta^6+\beta^3+3 \beta^3\left(\beta^2+\beta\right)=-p^3$
$\Rightarrow \quad q^2+q+3 q(-p)=-p^3 \quad$ [using Eq. (ii) $]$
$q^2+q-3 p q=-p^3$
$\Rightarrow \quad-p^3+3 p q=q^2+q$
$\Rightarrow \quad p\left(3 q-p^2\right)=q(q+1)$