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April 12th, 2015, 04:38 PM  #1 
Newbie Joined: Dec 2012 Posts: 12 Thanks: 0  Pythagorean and Quadratic equation Proof
Hello Everyone, I had a midterm in an mathematical econ class and we had to come up with a proof for the following question It was with regards to 2nd order difference equations, so Y(t+2) means Y subscript t+2: Show that for Y(t+2)+aY(t+1)+bY(t)=2, if 1+a+b=0 then it is always true that one of the roots of this equation is always equal to 1. The solution the professor came up with was: 1+a+b=0 then b=1a. therefore quadratic/characteristic equation shows that [a+sqrt(a^24(1)(1a))]/2(1) simplifies to [a+or  a+2]/2. this shows that at least one of the roots will always be 1. My proof was slightly different. if 1+a+b=0 then a+b=1, so for any values of a or b, using the quadratic equation [a+sqrt(a^24(b))]/2 this solves so that root 1=2/2 and root 2=some value/2. What I noticed was that using these parameters, the terms within the square root expression always follow the pythagorean theorem rule. Further more, the value inside the square root term is always +1 greater than the absolute value of the higher variable. However, I haven't been able to express the relationship. I'm hoping one of you who happen to be good at proofs can both flesh out my proof properly and explain the relationship to me mathematically. I couldn't find anything online. I am attaching a quick proof that I wrote, in case that helps. 

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equation, proof, pythagorean, quadratic 
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