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This question already has an answer here:

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?

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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

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

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

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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.

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