Solve x^4-16=0 in complex number system. Please explain. Thanks?

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Solve x^4-16=0 in complex number system. Please explain. Thanks?

2007-09-22 13:27:26 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

5 answers

Remember that (a²-b²)=(a+b)(a-b)

x^4 - 16 = 0
(x²)² - 4² = 0
(x² + 4)(x² - 4) = 0
((x² - (2i)²)(x² - 2²) = 0
(x+2i)(x-2i)(x+2)(x-2) = 0

x = {-2i, 2i, -2, 2}

2007-09-22 13:34:47 · answer #1 · answered by gudspeling 7 · 1⤊ 1⤋

x^4=16
take the 4th root of both sides
x= +/-2

2007-09-22 13:35:29 · answer #2 · answered by drewbear_99 5 · 0⤊ 2⤋

x^4 - 14 = 0
x^4 = 14
x = sqr(14) + 0 i

2007-09-22 13:32:51 · answer #3 · answered by CPUcate 6 · 0⤊ 3⤋

x^4 = 16
x = Â± 2

2007-09-22 22:15:37 · answer #4 · answered by Como 7 · 0⤊ 4⤋

(x^2 +4)(x^2 - 4) = 0
x^2 = -4
x = +/- 2i

OR
x^2 = 4
x = +/-2

2007-09-22 13:34:31 · answer #5 · answered by ccw 4 · 1⤊ 1⤋