1. ## Ratio...

If 2x^2 - 3xy + y^2 = 0, find the ratio of x:y.

2. Originally Posted by snigdha
If 2x^2 - 3xy + y^2 = 0, find the ratio of x:y.
The first step would be to divide through by y:

$\displaystyle \frac{2x^2}{y} - 3x + y = 0$

Then divide through by x:

$\displaystyle \frac{2x}{y} - 3 + \frac{y}{x} = 0$

Now, we move the last two terms to the other side of the equation:

$\displaystyle \frac{2x}{y} = 3 - \frac{y}{x}$

Then we divide through by 2:

$\displaystyle \frac{x}{y} = \frac{3}{2} - \frac{y}{2x}$

And there you go.

3. Another way: divide through immediately by $\displaystyle y^2$.

That gives you $\displaystyle 2(x/y)^2- 3(x/y)+ 1= 0$ a quadratic equation for x/y.

4. Originally Posted by snigdha
If 2x^2 - 3xy + y^2 = 0, find the ratio of x:y.
Divide through by y^2

2(x/y)^2 - 3(x/y) + 1 = 0

Solve for (x/y). Factoring works:

[2(x/y) - 1][(x/y) - 1] = 0

(x/y) = 1/2 or 1 ...x:y = 1:2 or 1:1

5. Hi snighda,
the previous two methods given by HallsofIvy and Diagonal
are the main variations.

$\displaystyle 2x^2-3xy+y^2=0$ may be converted to a quadratic in $\displaystyle \frac{x}{y}$

$\displaystyle \frac{2x^2-3xy+y^2}{y^2}=\frac{0}{y^2}$

$\displaystyle 2\left(\frac{x}{y}\right)^2-3\left(\frac{x}{y}\right)+1=0$

This is now of the form $\displaystyle aq^2+bq+c=0$

for which the solutions are $\displaystyle q=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Hence $\displaystyle 2\left(\frac{x}{y}\right)^2-3\left(\frac{x}{y}\right)+1=0$ for $\displaystyle \frac{x}{y}=\frac{3\pm\sqrt{9-4(2)}}{2(2)}=\frac{3\pm1}{4}$

You achieve the exact same result by writing the expression as a multiplication of factors

$\displaystyle 2x^2-3xy+y^2=(2x+u)(x+v)$

$\displaystyle uv=y^2,\ 2xv+ux=x(2u+v)=x(-3y)$

therefore $\displaystyle u=v=-y$

$\displaystyle 2x^2-3xy+y^2=0\ \Rightarrow\ (2x-y)(x-y)=0$

Two numbers multiplied give zero, hence either one of them can be zero.

which yields two solutions, x=y and 2x=y

from which the exact same ratios obtained from the quadratic may be written.

6. Originally Posted by HallsofIvy
Another way: divide through immediately by $\displaystyle y^2$.

That gives you $\displaystyle 2(x/y)^2- 3(x/y)+ 1= 0$ a quadratic equation for x/y.
Sorry, Halls. I'm certain that yours was not there when I posted my solution. I will have to check my time zone or whatever more closely.