2007-01-17 17:06:09 · 7 answers · asked by Car 1 in Science & Mathematics ➔ Mathematics
3x^2 =5x-2 subtract 5x-2 from each side
3x^2-5x+2=0
(3x-2)(x-1)=0
3x-2=0
3x=2
x=2/3
x-1=0
x=0
x=1, 2/3
2007-01-17 17:13:47 · answer #1 · answered by yupchagee 7 · 16⤊ 0⤋
3x^2 = 5x -2
3x^2 - 5x + 2 = 0
3x^2 -3x - 2 x + 2 = 0
3x(x - 1) -2(x - 1) = 0
(3x - 2) (x - 1) = 0
either, x = 2/3 or x = 1
2007-01-18 02:39:59 · answer #2 · answered by rishi 2 · 1⤊ 0⤋
Subtract 5x and add 2 from both sides:
3x^2-5x+2=0
Factor it:
(3x-2)(x-1)=0
Now, you solve:
3x-2=0
x-1=0
x=2/3 and 1
2007-01-18 01:30:33 · answer #3 · answered by Anonymous · 0⤊ 0⤋
3x^2 - 5x + 2 = 0
(3x -2)(x - 1) = 0
So either 3x - 2 = 0 ; 3x = 2; x = 2/3
or x - 1 = 0 ; x = 1
2007-01-18 01:13:53 · answer #4 · answered by ecolink 7 · 1⤊ 0⤋
3x^2=5x-2
=>3x^2-5x+2=0
=> 3x^2-3x-2x+2=0
=>3x(x-1)-2(x-1)=0
=>(x-1)(3x-2)=0
Hence either x-1=0 or 3x-2=0
kTherefore, x=1 or 2/3
2007-01-18 01:14:07 · answer #5 · answered by alpha 7 · 1⤊ 0⤋
3x^2=5x-2
3x^2-5x+2=0
3x^2-3x-2x+2=0
3x(x-1)-2(x+1)=0
(3x-2)(x+1)
so x= 2/3 , -1
2007-01-18 01:18:54 · answer #6 · answered by Anonymous · 0⤊ 1⤋
3x^2=5x-2
3x^2-5x+2=0
x=2/3
x=1
2007-01-18 01:22:44 · answer #7 · answered by supensa 6 · 1⤊ 1⤋