If $ax^2+bx+c=0$ has $2$ roots, one root is twice the other, find expression for $b$ $ax^2+bx+c=0$, has two roots with one being twice the other, so $x$ and $2x$
I need to find an expression for $b$ (in terms of $a$ and $c$)
I know $b = \dfrac{-ax^2-c}x$, but I don't know how to use the roots and how it would affect the answer
 A: If $x_1$ and $x_2=2x_1$ are the roots, then by Vieta's formulas $$x_1+x_2=3x_1=-\frac ba$$ and $$x_1x_2=2x_1^2=\frac ca\implies x_1=\sqrt{\frac c{2a}}$$ so subtituting into the first equation $$3\sqrt{\frac{c}{2a}}=-\frac ba\implies \frac{9c}{2a}=\frac{b^2}{a^2}\implies \boxed{b=3\sqrt{\frac{ac}2}}$$
A: Hint:
If $y,2y$ are the roots,
using Vieta's formula
$$y+2y=-\dfrac ba\iff y=?$$
$$y\cdot2y=\dfrac ca\iff y^2=?$$
Now eliminate $y$ by comparing the values of $y^2$
A: Without using the Viète-formulas:
$$x_1=2x_2$$
$$\frac{-b+\sqrt{b^2-4ac}}{2a}=2\frac{-b-\sqrt{b^2-4ac}}{2a}$$
$$\frac{-b+\sqrt{b^2-4ac}}{2}=-b+\sqrt{b^2-4ac}$$
$$\frac{b}{2}=-\frac{3}{2}\sqrt{b^2-4ac}$$
$$b^2=9(b^2-4ac)$$
$$8b^2=36ac$$
$$b=\pm \sqrt{\frac{36ac}{8}}$$
A: Let the roots be $y$ and $2y$. Then:
$$ay^2 + by + c = 0$$
and 
$$4ay^2 + 2by + c = 0$$
Multiplying the first by 4 and subtracting the two equations:
$$2by + 3c = 0$$
or
$$y = -\dfrac{3c}{2b}$$
Substituting this into the equation you had arrived at:
$$b = \dfrac{-ay^2-c}y$$
$$b = \dfrac{9ac^2+4b^2c}{6bc}$$
or
$$2b^2c = 9ac^2$$
$$2b^2 = 9ac$$
$$b^2 = \dfrac{9ac}2$$
$$\boxed{b = 3\sqrt{\dfrac{ac}{2}}}$$
