Question

x1 and x2 are solutions of x^2-x+1=0. What is the value of the sum (x1^2-x1)^1000 + (x2^2-x2)^1000 ?

Accepted Answer

x1 and x2 are given as the solutions of x^2 - x + 1 = 0
So if we substitute x with x1 we get x1^2 - x1 + 1 = 0
=> x1^2 - x1 = -1
Similarly with x2, x2^2 - x2 + 1 = 0
=> x2^2 - x2 = -1
The required sum is (x1^2-x1)^1000 + (x2^2-x2)^1000
=> -1^1000 + -1^1000
=> 1 + 1
=> 2
The required sum is 2.

Answer

We know the fact that the solution of an equation, substituted into equation, it cancels out the equation.
We'll substitute the solutions of the equation into the given equation:
x1^2 - x1 + 1 = 0
x1^2 - x1 = -1We'll raise to the power of 1000 both sides:
(x1^2-x1)^1000 = (-1)^1000
(x1^2-x1)^1000 = 1 (1)
x2^2 - x2 + 1 = 0
x2^2 - x2 = -1We'll raise to the power of 1000 both sides:
(x2^2-x2)^1000 = (-1)^1000
(x2^2-x2)^1000 = 1 (2)
We'll add (1) and (2):
(x1^2-x1)^1000 + (x2^2-x2)^1000 = 1 + 1
(x1^2-x1)^1000 + (x2^2-x2)^1000 = 2
The value of the sum = 2.

algebra1

x1 and x2 are solutions of x^2-x+1=0. What is the value of the sum (x1^2-x1)^1000 + (x2^2-x2)^1000 ?

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