Question 536504
Let r and 2r be the roots. By Vieta's formulas, the sum of the roots is -b/a and the product of the roots is c/a. We have


*[tex \LARGE 3r = -\frac{b}{a}]


*[tex \LARGE 2r^2 = \frac{c}{a}]


Squaring the first equation,


*[tex \LARGE 9r^2 = \frac{b^2}{a^2} \Rightarrow 2r^2 = \frac{2b^2}{9a^2}]


This is equal to c/a so we have


*[tex \LARGE \frac{2b^2}{9a^2} = \frac{c}{a}] This can be simplified and rearranged to obtain 2b^2 = 9ac.