2006-11-09 09:00:52 · 5 answers · asked by Dean F 1 in Science & Mathematics ➔ Mathematics
6x^3 - 7x = 0
x(6x^2-7)=0
x=0 is 1 root.
the ohter 2 can be found from
6x^2-7=0
6x^2=7
x^23=7/6
x=+/-√7/6) so
x=√(7/6)
x=-√(7/6) and
x=0 are the 3 roots.
2006-11-09 11:55:53 · answer #1 · answered by yupchagee 7 · 15⤊ 0⤋
you could first factor out an x, which would give you x(6x^2 - 7) = 0.
Then you would set x and 6x^2 - 7 to = 0 individually. That would give you x = 0 and x = plus or minus the square root of 7/6. (It is plus or minus because when you take the sqaure root of a number, you get two answers: a positive and negative value)
Hope that helps!
2006-11-09 17:06:12 · answer #2 · answered by Jenn 2 · 0⤊ 0⤋
undistribute the x
x(6x^2-7)=0
set both factors =0
x=0
and
6x^2-7=0 => x=(7/6)^.5
2006-11-09 17:09:19 · answer #3 · answered by rawfulcopter adfl;kasdjfl;kasdjf 3 · 0⤊ 0⤋
x(6x^2-7) = 0
x = 0
6x^2 - 7 = 0
6x^2 = 7
x^2 = 7/6
x = (7/6)^1/2
so x = 0, (7/6)^1/2
2006-11-09 17:03:03 · answer #4 · answered by disposable_hero_too 6 · 0⤊ 0⤋
Factor and use the zero product property.
2006-11-09 17:03:05 · answer #5 · answered by raz 5 · 0⤊ 0⤋