Maths - 4 root solution

5X^4 - 19X^3 + 22X^2 - 19X + 5 = 05x^2 - 19x + 22 - 19/x + 5/x^2 = 0 / x^25(x^2 + 1/x^2) - 19(x + 1/x) + 22 = 05[(x + 1/x)^2 - 2] - 19(x + 1/x) + 22 = 0Let x + 1/x = k ,5(k^2 - 2) - 19k + 22 = 05k^2 - 19k + 12 = 0(5k - 4)(k - 3) = 0k = 4/5 or k = 3x + 1/x = 4/5 or x + 1/x = 3x^2 - (4/5)x + 1 = 0 or x^2 - 3x + 1 = 0(no solutions) or x = [3 ± √(9 - 4)] / 2 x = (3 + √5)/2 or (3 - √5)/2

可以list埋條formular比我嗎? thanks a lot! ^.^

2010-09-21 03:16:54 補充：
really 雨中陽光... GREAT JOB!!!!! ^.^ 太棒了..!!!

2010-09-21 03:46:01 補充：
雨中陽光....太棒了!!!! ^.^

x=0.352(3 s.f.)
x=2.62(3 s.f.)