Question

Find the condition that one of roots of $a{x}^2+bx+c=0$ may be twice the other.

Answer 1

Let the roots of the equation $a{x}^2+bx+c=0$ are $\alpha ,2\alpha$

Sum of the roots $=\frac{-b}{a}$

$⟹\alpha +2\alpha =3\alpha =\frac{-b}{a}$

$⟹\alpha =\frac{-b}{3a}$

Product of roots $=\frac{c}{a}$

$\alpha \times 2\alpha =\frac{c}{a}$

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

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

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

$\frac{b^2}{9a^2}=\frac{c}{2a}$

$\frac{b^2}{9a}=\frac{c}{2}$

$\therefore 2b^2=9ac$

This is our required condition