1. ## [SOLVED] Hard Quadratic Equation

This is a straightforward math question to solve for R.

R^2(1/((Y^2)(X^2)))+R((2/(YX))-1)=1

I've spent a while trying to figure this out. Any help would be greatly appreciated! Thanks!

-Mike

2. Hello, Mike!

It seems to be a "simple" quadratic equation . . .

Solve for $\displaystyle R\!:\;\;R^2\left(\frac{1}{x^2y^2}\right) +R\left(\frac{2}{xy}-1\right)\;=\;1$

Multiply through by $\displaystyle x^2y^2\!:\;\;R^2 + xy(2-xy)R - x^2y^2 \;=\;0$

We have a quadratic equation with: .$\displaystyle a \,= \,1,\;\;b \,= \,xy(2-xy),\;\;c \,= \,\text{-}x^2y^2$

Then: .$\displaystyle x \;=\;\frac{-xy(2-xy) \pm \sqrt{[xy(2-xy)]^2 - 4(1)(\text{-}x^2y^2)}}{2(1)}$

. . . . . $\displaystyle x \;=\;\frac{xy(xy-2) \pm\sqrt{x^2y^2(2-xy)^2 + 4x^2y^2}}{2}$

. . . . . $\displaystyle x \;=\;\frac{xy(xy-2) \pm xy\sqrt{x^2y^2 - 4xy + 8}}{2}$

3. Alternatively just move the 1 over to the left and proceed with the quadratic formula
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