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mutatismutandis

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    fringe

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  1. Amaton thanks for the props! The ONLY thing I was able to find that remotely resembles the challenge was this: http://jliszka.github.io/2013/09/24/more-backwards-functions-unevaluating-polynomials.html And it was only posted a few days after the challenge was made live...and it only resembles the problem in the sense that a star is similar to an LED...sure they both make light, but they're operating via completely different processes.
  2. Fantastic, sorry I'm new didn't know how to operate latex on this site. :| Also, great catch on the [8] I had the 0= blah blah.... just missed it when I typed it into the text. Thanks for the puzzle and the kindness.
  3. here ya go...hopefully this looks ok... overkill.zip
  4. Ok....so I'm new here. bear with my lack of skill navigating the script for mathematical notation. Effectively, the goal is to get invert the polynomial so that we have x1, x2, y1, y2, and h in terms of a, b, c. This means that we'll eventually reach a point where we have one of the coefficients as a variable in a quadratic equation. let's begin: [1] y1=a*x1^2+b*x1+c [2]y2=a*x2^2+b*x2+c [3]c=h+(b^2/(4a)) subtract[2] from [1] to obtain [4] y1-y2=a(x1^2-x2^2)+b(x1-x2) for the sake of simplicity I now redefine the above equation as such [5] l=a*m+b*n, where l=y1-y2, m=x1^2-x2^2, and n=x1-x2; from here we need to isolate a variable so I choose to isolate a thus: [6] b=(l-a*m)/n I chose to substitute b into [1] to obtain [7] y1=a*x1^2+((l-a*m)/n)*x1+(h+((l-a*m)/n)^2/(4a))) now solve for a to obtain the quadratic function below: [8] y1= a^2*((m^2/(4n^2))-((m*x1)/n)+x1^2)+ a* (((l*x1)/n)-y1+h-((l*m)/(2n^2))) +(l^2/(4n^2)) when we solve this for a. we will obtain two solutions for a use each of those to back substitute into the y1 equation to obtain b and c. you will end up with two values for a, b, and c in terms of y1, y2, x1, x2, and h. I'm not going to type out the whole solution because I feel like the heart of the problem was to get this far and the rest of it is just mechanistic. So, do I get internet points?
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