If one root of $ax^2 + bx + c$ is double the other, then we must have $ 2b^2 = 9 ac $.

Question

If one root of $ax^2 + bx + c$ is double the other, then we must have $ 2b^2 = 9 ac $.

I cannot see how this is derived.

Answer 1

Let the two roots be $p$ and $2p$, then you have

$$ ax^2 + bx + c = a(x-p)(x-2p) = a(x^2-3px+2p^2) $$

Comparing coefficients gives $b = -3pa$ and $c = 2p^2a$, therefore

$$ p^2a^2 = \frac{b^2}{9} = \frac{ac}{2} $$

Answer 2

letting the roots be $r$ and $2r$, using Vieta's formulas we see that $\frac{-b}a = 3r$ and $\frac ca=2r^2$, so $2\left(\frac{-b}a\right)^2 = 18r^2 = \frac{9c}a$, i.e. $2b^2 = 9ac$.

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Bonus: if $2b^2 = 9ac$, then $b^2-4ac = b^2 - \dfrac{8b^2}9 = \dfrac19b^2$, so $x = \displaystyle \frac {-b \pm \sqrt{b^2-4ac}} {2a} = \frac {-b \pm \frac13b} {2a}$, so the two roots are $-\dfrac{b}{3a}$ and $-\dfrac{2b}{3a}$, so one is indeed double the other.