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