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Q.
The condition that one root of the equation $ax^{2}+bx+c=0$ may be square of the other, is
Complex Numbers and Quadratic Equations
Solution:
Let one root be $\alpha$ , then other root will be $\alpha^{2}$.
Given equation is $ax^{2} + bx + c = 0$
$\alpha$ is root of this equation
So, $a\alpha^{2} + b\alpha + c = 0$
Since sum of roots $=\alpha+\alpha^{2}=\frac{-b}{a}\, ...\left(i\right)$
Product of roots $=\alpha . \alpha^{2}=\frac{c}{a} \, ...\left(ii\right)$
Taking cube of equation $\left(i\right)$, we get
$\left(\alpha+\alpha^{2}\right)^{3}=\frac{-b^{3}}{a^{3}}$
$\Rightarrow \alpha^{3}+\left(\alpha^{2}\right)^{3}+3\alpha\alpha^{2}\left(\alpha+\alpha^{2}\right)=\frac{-b^{3}}{a^{3}}$
$\Rightarrow \,\frac{c}{a}+\left(\frac{c}{a}\right)^{2}+3\frac{c}{a}\left(\frac{-b}{a}\right)=\frac{-b^{3}}{a^{3}}$ (Using $(i)$ and $(ii)$)
$\Rightarrow \, \frac{c}{a}+\frac{c^{2}}{a^{2}}-\frac{3bc}{a^{2}}=\frac{-b^{3}}{a^{3}} $
or $ca^{2}+c^{2}a-3abc=-b^{3}$
or $a^{2}c+ac^{2}+b^{3}-3abc=0$.