# Determine the value(s) of the ratio $x:y$ if $2x^2-xy-3y^2 = 0$ [duplicate]

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?

## marked as duplicate by Daniel W. Farlow, graydad, Daniel, user91500, Claude LeiboviciJul 29 '15 at 5:51

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

$$\frac{x}{y} = \frac{3}{2} \textrm{ or } \frac{x}{y} = -1$$

You successfully found that $x = -y$ or $x = \frac{3}{2}y$. The question is asking for the ratio $\frac{x}{y}$. Hence, we may divide both sides of each equation by $y$ to obtain: $$\frac{x}{y} = -1 \text{ or } \frac{3}{2}$$

Hint:

$$2 x^2 - x y - 3 y^2 = 0$$

Let's say $x/y=k$. You are asked to find $k$. So you can write the equation above in the following form: $$2 x^2 - x^2k - 3x^2k^2 = x^2(2-k-3k^2)=0.$$ Solve this equation for $k$, which you have already done as I see.