Question

lf one root of the equation $a{x}^2+bx+c=0$ is the square of the other, then

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

Answer 1

The correct option is B $b^3+a{c}^2+a^{2}c=3abc$

Given equation $a{x}^2+bx+c$
Given that, one root of the equation is square of another.
So, lets assume $\alpha$ , ${\alpha }^2$ are roots of the given equation
We know that,
Sum of roots $=$ $\alpha +{\alpha }^2=\frac{-b}{a}$
Product of roots $=$ $\alpha \times {\alpha }^2=\frac{c}{a}$
$\alpha (1+\alpha )=\frac{-b}{a}⟶1$
${\alpha }^3=\frac{c}{a}⟶2$
Cubing equation (1) on both sides and substitute the value from equation (2).
${\alpha }^3{(1+\alpha )}^3=\frac{-b^3}{a^3}{\alpha }^3({\alpha }^3+1+3\alpha (1+\alpha \left)\right)=\frac{-b^3}{a^3}\frac{c}{a}(\frac{c}{a}+1+3\left(\frac{-b}{a}\right))=\frac{-b^3}{a^3}\frac{(c^2+ac-3bc)}{a^2}=\frac{-b^3}{a^3}a(c^2+ac-3bc)=-b^3{b}^3+a{c}^2+a^{2}c=3abc$