2006-09-16 12:59:45 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
A fifth degree equation in K. K = 0 is one solution to this equation. The others solve K^4 = 1/4, so the solutions are 1/sqrt(2), -1/sqrt(2), i/sqrt(2), and -i/sqrt(2). But this has nothing to do with E = MC^2.
2006-09-16 15:54:50 · answer #1 · answered by alnitaka 4 · 0⤊ 0⤋
K = 4K^5
4K^5 - K = 0 (subtract K from both sides)
K(4K^4 - 1) = 0 (factor out K)
K(2K^2 + 1)(2K^2 - 1) = 0 (factor 4K^4 - 1)
K=0 or 2K^2 + 1 = 0 or 2K^2 - 1 =0
solve for K...
K = 0
or
K = (sqrt 2)/2
or
K = - (sqrt)/2
2006-09-16 20:17:14 · answer #2 · answered by Giovanni McAdoo 4 · 0⤊ 0⤋
k=0
2006-09-16 20:01:57 · answer #3 · answered by Roger 4 · 0⤊ 1⤋
K = 4K^5
K = 0, is one choice
K = 4K^5
1 = 4K^4
1/4 = K^4
K = 1/4^-4
K = sqrt(sqrt(1/4))
K = sqrt( +/- 1/2)
K = +/- sqrt(2)/2 or +/- sqrt(2)i/2 or 0.
2006-09-16 20:03:28 · answer #4 · answered by Puzzling 7 · 1⤊ 0⤋
Hi. Only makes sense if K equals 0.
2006-09-16 20:02:15 · answer #5 · answered by Cirric 7 · 0⤊ 1⤋
The only value of K that would make this statement true is if K=0.
2006-09-16 20:27:50 · answer #6 · answered by whatthe 3 · 0⤊ 1⤋
an impossibility
2006-09-16 20:01:46 · answer #7 · answered by holden 4 · 0⤊ 1⤋
impossible
2006-09-16 20:02:50 · answer #8 · answered by a_blue_grey_mist 7 · 0⤊ 1⤋
Don't know,what?
2006-09-16 20:01:03 · answer #9 · answered by aries4272 4 · 0⤊ 1⤋