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 Score Degree Deduction (Posted on 2013-09-14)
Determine the total number of real solutions to the equation:
(x - 1)(x - 2)…..(x - 19)(x - 20) + 1 = 0, whenever 9 < x < 13.

Prove that there are no others.

 No Solution Yet Submitted by K Sengupta No Rating

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There are 4 solutions to the problem posed.
If we ignore the +1 the polynomial clearly has a root at each of 1...20
Noting its end behavior we know it has local minima in each of (1,2),(3,4)...(19,20).
If you shift the polynomial up 1 unit, the roots will either:
1). Move inward a bit.
2). Move inward and become a double root.
3). Become non-real.

Because each of the minima between the roots are so low (-thousands at least) option 1). Occurs each time.

So there are four solutions:  two each on (9,10) and (11,12).

 Posted by Jer on 2013-09-14 13:14:44

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