Question

If one root of the equation $a{x}^2+bx+c=0$ be the square of the other, then the value of $b^3+a^{2}c+a{c}^2$ is

• A. $3abc$
• B. $abc$
• C. $6abc$
• D. $\frac{abc}{3}$

Answer 1

The correct option is A $3abc$

Let one roots be $\alpha$ then other root is ${\alpha }^2$

$\therefore$ Sum of roots $\Rightarrow \alpha +{\alpha }^2=-\frac{b}{a}$ ...(1)

and $\alpha \times {\alpha }^2=\frac{c}{a}\Rightarrow {\alpha }^3=\frac{c}{a}$ ...(2)

From (1), we have

$\alpha (\alpha +1)=\frac{-b}{a}$

$\Rightarrow {\left(\alpha (\alpha +1)\right)}^3={\left(\frac{-b}{a}\right)}^3$

$\Rightarrow {\alpha }^3({\alpha }^3+3{\alpha }^2+3\alpha +1)=\frac{-b^3}{a^3}$

$\Rightarrow \frac{c}{a}(\frac{c}{a}+3\left(\frac{-b}{a}\right)+1)=\frac{-b^3}{a^3}$

$\Rightarrow \frac{c^2}{a^2}-\frac{3bc}{a^2}+\frac{c}{a}=\frac{-b^3}{a^3}$

$\Rightarrow {ac}^2-3abc+a^{2}c=-b^3$

$\Rightarrow b^3+{ac}^2+a^{2}c=3abc$