Question

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

Let roots of the equation $a{x}^2+bx+c=0$ be $\alpha$ and ${\alpha }^2$.

Then,

$\alpha +{\alpha }^2=-\frac{b}{a}$

$\alpha .{\alpha }^2={\alpha }^3=\frac{c}{a}$

$\alpha =(\frac{c}{a}{)}^{1/3}$

Thus, $(\frac{c}{a}{)}^{1/3}+(\frac{c}{a}{)}^{2/3}=-\frac{b}{a}$

Cubing both sides, we get,

$\frac{c}{a}+(\frac{c}{a}{)}^2+3.(\frac{c}{a}{)}^{1/3}\left(\frac{c}{a}{)}^{2/3}\right[(\frac{c}{a}{)}^{1/3}+(\frac{c}{a}{)}^{2/3}]=-\frac{b^3}{a^3}$

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

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

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

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