ok im in algebra 2 and i have this problem
3x^2-5X-2=0 that is the problem so i did this i dont know if its right
3X(3X^2 - 6X) 1(1X-2)
(x-2) (X-2)
I dont think this is right because i thought one had to be positive and one negative ...i may be wrong i was just wondering if someone could help me.. thanks
2006-11-06 08:56:21 · 3 answers · asked by Shirley B 1 in Science & Mathematics ➔ Mathematics
3x^2-5X-2=0
(3x+1)(x-2)=0
3x+1=0
x=-1/3
x-2=0
x=2
2006-11-06 08:59:09 · answer #1 · answered by yupchagee 7 · 15⤊ 0⤋
I don't understand your work.
3x^2 - 5X - 2=0
You expect the factors to look like
(ax + b)(cx + d)
In order to get 3 as the coefficient of the x^2 term, the choices for a & c are 1 & 3.
In order to get 2 as the constant term, the choices for b & d are 1 & 2.
By trial and error, try to find a, b, c, d to get - 5 as the coefficient of the x term.
And this doesn't seem to be possible. If you are trying to solve a quadratic, you will have to use the quadratic formula.
2006-11-06 17:08:13 · answer #2 · answered by kindricko 7 · 0⤊ 0⤋
factors are
(3x+1) ( x -2 )
By trial and error or by quadractic formula
2006-11-06 17:17:59 · answer #3 · answered by Jim C 3 · 0⤊ 0⤋