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11 months 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
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11 months ago
```							arun has not proved that no of real roots shall be 1 or 2 when n is odd or even.x^n= 1take modulus both sides|x^n| = |1|or |x|^n = 1or |x|^n – 1 = 0(|x| – 1)[|x|^(n-1) + |x|^(n-2) + …....... + |x|^1 + 1] = 0clearly |x|^(n-1) + |x|^(n-2) + …....... + |x|^1 + 1 can never be 0 since it always greater than 0.Hence |x| – 1 = 0or |x| = 1or x= 1, – 1now, 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 :))
```
11 months ago
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