Question

If one root of the quadratic equation $a{x}^2+bx+c=0$ is the square of the other, then $b^3+a^{2}c+a{c}^2=3abc$
Say yes or no.

• A. Yes
• B. No
• C. Ambiguous
• D. Data insufficient

Answer 1

The correct option is A Yes

Let one root of $a{x}^2+bx+c=0$ be $\alpha$ and other be ${\alpha }^2$
then, $\alpha +{\alpha }^2=\frac{-b}{a}$
${\alpha }^3=\frac{c}{a}$
or $\alpha =(\frac{c}{a}{)}^{\left(\frac{1}{3}\right)}$
Now put the value of $\alpha$ in $\alpha +{\alpha }^2=\frac{-b}{a}$
${\frac{c}{a}}^{\frac{1}{3}}+{\frac{c}{a}}^{\frac{2}{3}}=\frac{-b}{a}$
Cubing both sides:

$(\frac{c}{a}{)}^{\left(\frac{3}{3}\right)}+(\frac{c}{a}{)}^{\left(\frac{6}{3}\right)}+3(\frac{c}{a}{)}^{\left(\frac{1}{3}\right)}\times (\frac{c}{a}{)}^{\left(\frac{2}{3}\right)}\left(\right(\frac{c}{a}{)}^{\left(\frac{1}{3}\right)}+\left(\frac{c}{a}{)}^{\left(\frac{2}{3}\right)}\right)=(\frac{-b}{a}{)}^3$

$\frac{c}{a}+\frac{c^2}{a^2}+3\frac{c}{a}\left(\frac{-b}{a}\right)=\frac{-b^3}{a^3}$

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