Question 1135484: What three numbers, other than the number 1, when raised to the 4th power equals 1?
Found 2 solutions by ikleyn, jim_thompson5910:
Answer by ikleyn(52810)   (Show Source):
Answer by jim_thompson5910(35256)  (Show Source):
You can put this solution on YOUR website!
x = some unknown number
x^4 = raising that unknown number to the fourth power
x^4 = 1 is the equation we want to solve
We're told that x = 1 is one of the four solutions
x^4 = 1
x^4 - 1 = 0
(x^2)^2 - 1 = 0
(x^2-1)(x^2+1) = 0 ... difference of squares rule
(x-1)(x+1)(x^2+1) = 0 ... difference of squares rule
Set each factor equal to zero, through the zero product property, then solve for x
x-1 = 0 becomes x = 1, which is the solution mentioned earlier
x+1 = 0 becomes x = -1 which is another solution
x^2+1 = 0 becomes x^2 = -1 and that solves to x = i or x = -i where i = sqrt(-1)
"sqrt" is shorthand for "square root". The term sqrt(-1) is imaginary because we cannot apply the square root to a negative number, but this is defined to help solve equations such as x^2 = -1
The four roots of x^4 = 1 are:
x = 1 or x = -1
x = i or x = -i
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Checking the answers:
Let's check x = 1
x^4 = 1
1^4 = 1 ... plug in x = 1
1 = 1 ... true equation, solution is confirmed
Now check x = -1
x^4 = 1
(-1)^4 = 1 ... plug in x = -1
1 = 1 ... true equation, solution is confirmed
Now check x = i
x^4 = 1
(i)^4 = 1 ... plug in x = i
(i^2)^2 = 1
(-1)^2 = 1
1 = 1 ... true equation, solution is confirmed
Now check x = -i
x^4 = 1
(-i)^4 = 1 ... plug in x = -i
((-i)^2)^2 = 1
(i^2)^2 = 1
(-1)^2 = 1
1 = 1 ... true equation, solution is confirmed
All four solutions have been confirmed.
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