The zeroes of $ax² + bx + c = 0$ are in the ratio $1:2$ (assume the coefficients to be non-zero). Which of these is one of the zeroes?

Since the zeroes are in ratio $1:2$, let them be $k$ and $2k$.

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For any quadratic $ax^2 + bx + c = 0$:

Sum of zeroes $= -\dfrac{b}{a}$

Product of zeroes $= \dfrac{c}{a}$

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Sum of zeroes:

$k + 2k = 3k$

$3k = -\dfrac{b}{a}$

$k = -\dfrac{b}{3a}$

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Product of zeroes:

$k \times 2k = 2k^2$

$2k^2 = \dfrac{c}{a}$

$k^2 = \dfrac{c}{2a}$

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Substituting $k = -\dfrac{b}{3a}$ into $k^2 = \dfrac{c}{2a}$:

$\left(-\dfrac{b}{3a}\right)^2 = \dfrac{c}{2a}$

$\dfrac{b^2}{9a^2} = \dfrac{c}{2a}$

$2ab^2 = 9a^2c$

Since $a \neq 0$:

$2b^2 = 9ac$

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The two zeroes are:

First zero: $k = -\dfrac{b}{3a}$

Second zero: $2k = -\dfrac{2b}{3a}$

Therefore, one of the zeroes is $-\dfrac{b}{3a}$ or $-\dfrac{2b}{3a}$.