Let one root of the equation \(ax^2+bx+c=0\) be \(α\) and the other root be \(2α\).
\(∴ α+2α=-\cfrac{b}{a}\)
or, \(3α=-\cfrac{b}{a}\)
or, \(α=-\cfrac{b}{3a}\)
Again, \(α.2α=\cfrac{c}{a}\)
or, \(2α^2=\cfrac{c}{a}\)
or, \(2×(-\cfrac{b}{3a})^2=\cfrac{c}{a}\)
or, \(\cfrac{2b^2}{9a^2}=\cfrac{c}{a}\)
or, \(2ab^2=9a^2 c\)
or, \(2b^2=9ac\) (Proved).