I'm confused! I got the wrong answer cuz it's not one of the choices...
3w^2=7w+20
2007-05-22 17:58:56 · 4 answers · asked by jamie68117 3 in Science & Mathematics ➔ Mathematics
The instructions say: Solve the following equations over the set of complex numbers.
2007-05-22 18:00:19 · update #1
easiest to solve by factoring
3w² - 7w - 20 = 0
(3w + 5 ) ( w - 4 ) = 0
w = - 5/3 OR w = 4
in this case, all solutions are real and rational.
=]
..
2007-05-22 18:02:00 · answer #1 · answered by Anonymous · 0⤊ 1⤋
Get all terms on one side....
3w^2 - 7w -20 = 0
This factors: (3 w + 5 )(w - 4) = 0
Then w = 4 and w = - 5/3
[since b^2 - 4 ac is positive, the answers won't be complex numbers]
2007-05-23 01:06:29 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋
First set this equal to zero
3w^2 - 7w -20 = 0
now factor
(3w + 5)(w - 4) = 0
then
3w + 5 = 0 3w = -5 w = -5/3
w - 4 = 0 w = 4
so the solution set is -5/3 and 4
2007-05-23 01:05:31 · answer #3 · answered by msmthtchr 3 · 0⤊ 0⤋
3x^2=7x+20
-3x^2
-3x^2 + 7x + 20 = 0
http://www.math.com/students/calculators/source/quadratic.htm
x = -3/2 and 4
2007-05-23 01:04:57 · answer #4 · answered by cosmichorizon 3 · 0⤊ 0⤋