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Let `f(x) = x^2+bx+c` and `g(x) = x^2+qx+r` , where `b, c, q, r in RR` and `c != r` Let...
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`g(x) = x^2+qx+r = 0` with roots `alpha` and `beta`
So we can say;
`alpha+beta = -q`
`alphaxxbeta = r`
So the required answer is obtained.
Posted by jeew-m on September 25, 2013 at 7:54 AM (Answer #1)
High School Teacher
There's kind of a cool way to solve it with less factoring, which is nice for people like me who aren't the quickest at finding factorizations.
Note that `f(x)-g(x)=(b-q)x+(c-r).` Then since `g(alpha)=g(beta)=0,` we have
So you can already see where the `b-q`and `c-r` terms come from in the answer (actually those terms were what led me to try subtracting the functions in the first place). Now multiply these (you have to do a little bit of factoring at times too) and use the fact that `alpha*beta=r` and `alpha+beta=-q,` and the answer just pops out. Since the question was already solved though, I'll leave the straightforward details of this to you.
Posted by degeneratecircle on September 25, 2013 at 11:31 PM (Answer #2)
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