Determine the value(s) of the ratio $x:y$ if $2x^2-xy-3y^2 = 0$ I was asked to determine the value(s) of the ratio $x:y$ if $2x^2-xy-3y^2 = 0$. I didn't know what this meant, so I just solved the equation to get $x = -y$ or $x = \frac{3}{2}y$. What does the question mean?

As $x:y$ can be written as $\displaystyle \frac{x}{y}$, then
$$ 2 x^2 - x y - 3 y^2 = 0 $$
can be written as
$$ \frac{ 2 x^2 - x y - 3 y^2 = 0 }{ y^2 } = 0 $$,
If $y \ne 0$, thus
$$ 2 \left( \frac{x}{y} \right)^2 - \left( \frac{x}{y} \right) - 3 = 0 $$
Or
$$ \left( \frac{x}{y} \right)^2 - \frac{1}{2} \left( \frac{x}{y} \right) - \frac{3}{2} = 0 $$
Then
$$ \left[ \left( \frac{x}{y} \right) - \frac{1}{4} \right]^2 - \frac{1}{16} - \frac{3}{2} = 0 $$
so
$$ \left[ \left( \frac{x}{y} \right) - \frac{1}{4} \right]^2  = \frac{25}{16} $$
then
$$\frac{x}{y}  - \frac{1}{4} = \pm \frac{5}{4}$$
or
$$\frac{x}{y}  = \frac{1}{4} \pm \frac{5}{4}$$
Thus
$$\frac{x}{y} = \frac{3}{2} \textrm{ or } \frac{x}{y} = -1 $$
