2007-08-13 12:07:57 · 5 answers · asked by rachel z 1 in Science & Mathematics ➔ Mathematics
(x^4 - 1)/(x^3) = 0
multiply both sides by x^3.
x^4 - 1 = 0
x^4 = 1
x = +/- 1
2007-08-13 12:13:10 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Provided x is not 0 (in which case the "equation" is meaningless!), multiply by x^3.
Then x^4 - 1 = 0.
Now factorize: (x^2 - 1)(x^2 + 1) = 0, that is
(x - 1)(x + 1)(x^2 + 1) = 0.
Thus there are 4 roots, x = +/- 1 and +/- i, where i is the square root of -1. QED
Live long and prosper.
2007-08-13 12:17:42 · answer #2 · answered by Dr Spock 6 · 0⤊ 0⤋
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2007-08-13 12:12:35 · answer #3 · answered by Anonymous · 0⤊ 1⤋
(x^4 -1) / (x^3)=0
x^4-1 =0 <-- multiply both sides by x^3
x^4 = 1
x = 1,i,-1,-i
2007-08-13 12:21:52 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋
(x^4 -1) / (x^3) = 0
Multiplying both sides by x^3
x^4 - 1 = 0
Factorizing
(x - 1)(x +1)(x^2 + 1) = 0
So
x = 1
x = - 1
x = +- i where i = imaginary number = sqrt of (-1)
2007-08-13 12:19:30 · answer #5 · answered by Sheen 4 · 0⤊ 0⤋