Question

The condition that one root of the equation $a{x}^2+bx+c=0$ may be square of the other, is

• A. $a^{2}c+a{c}^2+b^3-3abc=0$
• B. $a^2{c}^2+a{c}^2+b^2+3abc=0$
• C. $a{c}^2+ac-b^3-3abc=0$
• D. $a^{2}c+a{c}^2-b^3-3abc=0$

Answer 1

Answer: (A)

Let one root be $\alpha$ , then other root will be ${\alpha }^2$.
Given equation is $a{x}^2+bx+c=0$
$\alpha$ 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 $c{a}^2+c^{2}a-3abc=-b^3$
or $a^{2}c+a{c}^2+b^3-3abc=0$.