Question

If one root of the equation $a{x}^2+bx+c=0$ the square of the other, then $a(c-b{)}^3=cX,$ where $X$ is

• A. $a^3-b^3$
• B. $a^3+b^3$
• C. $(a-b{)}^3$
• D. None of these

Answer 1

The correct option is C

$(a-b{)}^3$

If one root is square of other of the equation $a{x}^2+bx+c=0,$ then product of roots $=\frac{c}{a}$ and
sum of roots $=-\frac{b}{a}$.
If we assume the root of the equation as $\alpha$ and ${\alpha }^2$ then ${\alpha }^3=\frac{c}{a}$ and
$\alpha$ + ${\alpha }^2=-\frac{b}{a}$
On simultaneously solving these equations and cubing we will get,
$b^3+a{c}^2+a^{2}c=3abc$

Which can be written in the form $a(c–b{)}^3=c(a–b{)}^3$

Alternate solution
Trick: Let roots be 2 and 4, then the equation is $x^2$ – 6x + 8 = 0. Here obviously

$X=\frac{a(c-b{)}^2}{c}=\frac{1(14{)}^3}{8}=\frac{14}{2}\times \frac{14}{2}\times \frac{14}{2}=7^3$

Which is given by $(a-b{)}^3=7^3$