Let one root of the quadratic equation \(ax^2 + bx + c = 0\) be \(\alpha\), and the other root be \(2\alpha\).
\(\therefore \alpha + 2\alpha = -\cfrac{b}{a}\) Or, \(3\alpha = -\cfrac{b}{a}\) Or, \(\alpha = -\cfrac{b}{3a}\)
Again, \(\alpha \cdot 2\alpha = \cfrac{c}{a}\) Or, \(2\alpha^2 = \cfrac{c}{a}\) Or, \(2 \times \left(-\cfrac{b}{3a}\right)^2 = \cfrac{c}{a}\) Or, \(\cfrac{2b^2}{9a^2} = \cfrac{c}{a}\) Or, \(2ab^2 = 9a^2c\) Or, \(2b^2 = 9ac\) (Proved)