If $ax^2+bx+c=0$ has $2$ roots, one root is twice the other, find expression for $b$

Question

$ax^2+bx+c=0$, 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

Answer 1

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 $y^2$

Answer 2

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 $$3\sqrt{\frac{c}{2a}}=-\frac ba\implies \frac{9c}{2a}=\frac{b^2}{a^2}\implies \boxed{b=3\sqrt{\frac{ac}2}}$$

Answer 3

Without using the Viète-formulas: $$x_1=2x_2$$ $$\frac{-b+\sqrt{b^2-4ac}}{2a}=2\frac{-b-\sqrt{b^2-4ac}}{2a}$$ $$\frac{-b+\sqrt{b^2-4ac}}{2}=-b+\sqrt{b^2-4ac}$$ $$\frac{b}{2}=-\frac{3}{2}\sqrt{b^2-4ac}$$ $$b^2=9(b^2-4ac)$$ $$8b^2=36ac$$ $$b=\pm \sqrt{\frac{36ac}{8}}$$

Answer 4

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

$$2b^2c = 9ac^2$$

$$2b^2 = 9ac$$

$$b^2 = \dfrac{9ac}2$$

$$\boxed{b = 3\sqrt{\dfrac{ac}{2}}}$$