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Part (b) of the upper question. Please help with the solution of the aforementioned question.

Part (b) of the upper question.
Please help with the solution of the aforementioned question.

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Grade:12

2 Answers

Arun
25750 Points
4 years ago
Dear student
 
xⁿ -1 = 0
 
Now there will be n roots as this polynomial of degree n.
 
Sum of roots = - (coefficient of x n-1 ) / coefficient of xⁿ
 
 = 0
 
Hence proved
Aditya Gupta
2081 Points
4 years ago
arun has not proved that no of real roots shall be 1 or 2 when n is odd or even.
x^n= 1
take modulus both sides
|x^n| = |1|
or |x|^n = 1
or |x|^n – 1 = 0
(|x| – 1)[|x|^(n-1) + |x|^(n-2) + …....... + |x|^1 + 1] = 0
clearly |x|^(n-1) + |x|^(n-2) + …....... + |x|^1 + 1 can never be 0 since it always greater than 0.
Hence |x| – 1 = 0
or |x| = 1
or x= 1, – 1
now, coming back to the original eqn:
x^n = 1, we have 2 cases:
  1. n= odd, if we put x= 1 in the eqn x^n= 1, it is satisfied. but if we put x= – 1, then x^n= ( – 1)^n= – 1 (since n is odd). so, there is only one real soln x= 1
  2. n= even, if we put x= 1 in the eqn x^n= 1, it is satisfied. if we put x= – 1, then x^n= 1 (since n is even). so, there are 2 real solns x= 1 and – 1.
​kindly approve :))
 

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