Real roots again?
Consider the polynomial P n ( x ) = x n + x n − 1 + x n − 2 + … + x + 1
Find the maximum no.of distinct real roots it can have if n is a positive integer strictly less than 1000.
The answer is 1.
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2 solutions
Hi, would you please explain to me how did you decide to define G n ( x ) the way you did ?
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I guess because the roots of r = 0 ∑ n − 1 x r are all the n th roots of unity except 1 .
Or, you could simply apply the "formula" for the sum of the terms of a geometric sequence and see that x − 1 appears in the denominator. Multiply to get rid of it.
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Thanks bro, nice observation . Considering that you are just two years my junior, your mind is quite sharp :)
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@Honey Singh – Lol I'm actually 17 :P
@Honey Singh – Are you the singer Honey Singh ? I'm asking this since your age is also 31
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@A Former Brilliant Member – I wish I weere , but no i am his fan . of 31 years too
from where did u get this property mentioned ion 1st line?
Define another polynomial G n ( x ) = ( x − 1 ) P n ( x )
So, G n ( x ) = x n + 1 − 1
If G n ( x ) = 0 , then x n + 1 = 1
If n + 1 is even, then we get two distinct real solutions, − 1 and 1 .
If n + 1 is odd, then we get only one real solution, 1 .
In each case one solution comes from the factor ( x − 1 ) .
So, P n ( x ) can have only 1 real solution.