2007-12-18 03:24:17 · 6 answers · asked by RivalMikhail 1 in Science & Mathematics ➔ Mathematics
Real Roots:
x = +/- quarticroot 8
FCH.
2007-12-18 03:27:26 · answer #1 · answered by Anonymous · 0⤊ 0⤋
- 8^(1/4) and 8^(1/4) are the real solutions. The non-real solutions are - 8^(1/4)*i and 8^(1/4)*i if you also want those.
2007-12-18 11:28:46 · answer #2 · answered by ben e 7 · 0⤊ 0⤋
2^1/2
2007-12-18 11:27:24 · answer #3 · answered by Anonymous · 0⤊ 2⤋
Since it is a polynomial of degree 4 it has four roots: (2 real and 2 imaginary)
x^4 = 8
take sqaure root of both sides
x^2 = +- sqrt(8)
take square root of both sides:
x = +-sqrt(+- sqrt(8))
answers:
sqrt(sqrt(8)) = 8^(1/4)
sqrt(-sqrt(8)) = i8^(1/4)
-sqrt(sqrt(8)) = -8^(1/4)
-sqrt(-sqrt(8)) = -i8^(1/4)
2007-12-18 11:33:23 · answer #4 · answered by KEYNARDO 5 · 0⤊ 0⤋
x^4-8=0
(x^2-â8)(x^2+â8)=0
(x- 8^¼)(x+ 8^¼) (x^2+â8) =0
x= +/-8^¼
[in real numbers only +/- 8^¼
but in imaginary numbers also +/- i 8^¼ ]
2007-12-18 11:32:13 · answer #5 · answered by mbdwy 5 · 0⤊ 0⤋
â2 â2 â2 â2 = 8
(-â2) (-â2) (-â2) (-â2) = 8
Thus x = 屉2
2007-12-18 14:25:08 · answer #6 · answered by Como 7 · 4⤊ 0⤋