Question
find roots of w^(4)+2^(3)+3^(2)+4w-52=0
Answers
Reiny
ok, I read
w^4 + 8 + 9 + 4w - 52 = 0
w^4 + 4w -35=0
No easy way to solve this,
Wolfram shows two real solutions
w = -2.5955 and w = 2.25747
http://www.wolframalpha.com/input/?i=w%5E4+%2B+4w+-35%3D0
notice you also have 2 complex roots.
w^4 + 8 + 9 + 4w - 52 = 0
w^4 + 4w -35=0
No easy way to solve this,
Wolfram shows two real solutions
w = -2.5955 and w = 2.25747
http://www.wolframalpha.com/input/?i=w%5E4+%2B+4w+-35%3D0
notice you also have 2 complex roots.
Steve
Assuming the usual careless typing of posters, I read it as
w^4+2w^3+3w^2+4w-52=0
a little synthetic division yields
(w-2)(w^3 + 4w^2 + 11w + 26)
and no easy roots thereafter
(one irrational, two complex)
w^4+2w^3+3w^2+4w-52=0
a little synthetic division yields
(w-2)(w^3 + 4w^2 + 11w + 26)
and no easy roots thereafter
(one irrational, two complex)