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This problem might be one of the hardest polynomial problems, I have encountered. Any help you can offer would be appreciated.

Given 26 equations in the form (x^2)+ax+c. The coefficients of x in the 26 respective equations have values: -1, -3, -5, -7 ...-51
If the roots of each equation are consecutive integers, find the sum of the squares of the 52 roots of the 26 equations. (2 roots per equation)

Answer
Questioner:   Isaac
Country:   Illinois, United States
Category:   Advanced Math
Private:   No  <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<< changed
Subject:   difficult polynomial problem
Question:   This problem might be one of the hardest polynomial problems, I have encountered. Any help you can offer would be appreciated.

Given 26 equations in the form (x^2)+ax+c. The coefficients
of x in the 26 respective equations have values: -1, -3, -5,
-7 ...-51
If the roots of each equation are consecutive integers, find
the sum of the squares of the 52 roots of the 26 equations.

(2 roots per equation)
............................
Have you studied quadratic equations?  Then you learned that

in the equation:

x^2 + ax + c  

the sum of the roots is equal to  -a.  

Thus if the equation is:

x^2 - 5x + c = 0.   <<<<< don't leave out the  "= 0"

then the roots add up to 5.  If that is so, then you have a

basic algebra problem:

Find two consecutive integers whose sum is 5.

Your little sister could do that one.

So you have:

Find two consecutive integers whose sum is  1.
Find two consecutive integers whose sum is  3.
Find two consecutive integers whose sum is  5.
......
Find two consecutive integers whose sum is 51.

I think you can take it from there.

P.S. You will need the (well-known) formula for the sum of
the squares of the first 'n' integers.

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