If x² + 4y² = 4xy, find x : y.

Given,

x² + 4y² = 4xy

Dividing both sides by xy we get,

$\Rightarrow \dfrac{x^2 + 4y^2}{xy} = \dfrac{4xy}{xy} \\[1em] \Rightarrow \dfrac{x}{y} + 4\dfrac{y}{x} = 4$

Let $\dfrac{x}{y}$ = t

$\Rightarrow t + 4\dfrac{1}{t} = 4 \\[1em] \Rightarrow \dfrac{t^2 + 4}{t} = 4 \\[1em] \Rightarrow t^2 + 4 = 4t \\[1em] \Rightarrow t^2 - 4t + 4 = 0 \\[1em] \Rightarrow t^2 - 2t - 2t + 4 = 0 \\[1em] \Rightarrow t(t - 2) - 2(t - 2) = 0 \\[1em] \Rightarrow (t - 2)(t - 2) = 0 \\[1em] \Rightarrow t - 2 = 0 \text{ or } t - 2 = 0 \\[1em] \Rightarrow t = 2 \\[1em] \Rightarrow \dfrac{x}{y} = 2.$

Hence, x : y = 2 : 1.