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

Grade 12

2 Answers

Last Activity: 5 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

Last Activity: 5 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:

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
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 :))