P ( x ) = x 6 + ( x − r ) 6 + ( x − 2 r ) 6 − 3 r The polynomial above has exactly two real zeros. Find the value of r such that the sum of the real zeros is 1 .
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Thats my way :)
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Yes I agree that i first took the vision of the problem straight from your solution . But the concept involved here is very old though. This doesn't belong to a particular individual.
P.S. BTW your solution was really awesome.
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Indeed it belongs only to the mathematicians , i only meant that this perhaps partly came from the solution i posted :)
As soon as I put x-r=t. I knew i would find this comment here :)
If r is complex the line x = r does not exist.
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I asked to find the value of r . Doesn't it imply that it is real?
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For a complete solution all cases should be considered.
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@Joel Tan – Thanks. See this problem.
@Joel Tan – We don't need any line. The fact that P ( r − x ) = P ( r + x ) is sufficient to get the result. Even when r is an imaginary number. That was for the sake of visual understanding.
https://brilliant.org/problems/find-the-areaonly-your-logic-can-help-you/?group=3UHxOzwinQpA&ref_id=384997
PLEASE TRY TO DO THIS AWSOME PROBLEM TOO..post a solution if you get......................i am waiting for an awesome solution that i made while creating this problem
Solved by graphing trying different values of r.
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We observe that P ( r + x ) = P ( r − x ) for all x . That means the graph of the polynomial is symmetrical with respect to the line x = r . Further among 2 real zeros, if one zero is ( r + S ) the corresponding other would be ( r − S ) . That means the sum of the 2 real zeros will be 2 r . 2 r = 1 ⇒ r = . 5