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can help with algebraic approach, yet discriminant solution is tedious

I am using FOIL for p^2-kp-28=0
possible solutions (p-28)(p+1)=0, (p-14)(p+2)=0, (p-7)(p+4)=0
in all solutions we get the value -28 for independent argument (C) and different values for k such as -28+1=-27, -14+2=-12, -7+4=-3. Since k is positive and our quadratics contains -1*k we obtain the sum of all possible values for k as 27+2+3=42

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p^2-kp-28=0

using Vieta's theorem (http://en.wikipedia.org/wiki/Vieta%27s_formulas)

p1*p2=-28
p1+p2=-(-k)

p1 p2 can be (28; -1) ; (14;-2); (7;-4)

28-1+14-2+7-4=42

p.s. I forgot this formula, but few days ago I did math of my nephew, and was pleased to see such a useful theorem )))) verdict- sometimes it is good to help kids )))

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@knight, why in your post you seek discriminant solution? Is this something related to graphical solution of equation?

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