Question

One root of the equation $a{x}^2+bx+c=0$ is square of the other if

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

Answer 1

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

Let the roots of the equation $a{x}^2+bx+c=0$ be $p,p^2$
then, Sum of the roots = $p+p^2=-\frac{b}{a}$
Product of theroots = $p^3=\frac{c}{a}$
$\Rightarrow p=(\frac{c}{a}{)}^{\frac{1}{3}}$
Now ,$p+p^2=-\frac{b}{a}$
$(\frac{c}{a}{)}^{\frac{1}{3}}+(\frac{c}{a}{)}^{\frac{2}{3}}=-\frac{b}{a}$
Cube both sides.
$\frac{c}{a}+(\frac{c}{a}{)}^2+3(\frac{c}{a}\left)\right(\frac{-b}{a})=-(\frac{b}{a}{)}^3$
$a^{2}c+a{c}^2-3acb=-b^3$
$a^{2}c+b^3+a{c}^2=3abc$