2007-05-29 14:29:14 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
3x^2 + 5x - 8 = 0
(3x + 8)(x - 1) = 0
So x = 1 is the postive root
2007-05-29 14:33:32 · answer #1 · answered by peateargryfin 5 · 0⤊ 0⤋
Step 1: Subtract 8 from both sides of the equation.
3x^2 + 5x = 8
3x^2 + 5x - 8 = 0
Step 2: Find factors of (-8)(3) = -24 that add to give 5. These are 6 and -1.
Step 3: Use a factoring technique such as the box/grid method, the quadratic formula, trial and error, etc., to figure out what coefficients you need in front of the x's.
If we do this, we see that (x - 1)(3x + 8) satisfies this equation.
This means x = 1 and x = -8 are solutions to this equation. x = 1 is the only positive solution/root.
2007-05-29 21:37:30 · answer #2 · answered by JoeSchmo5819 4 · 0⤊ 1⤋
3x^2+5x-8=0
(3x+8)(x-1)=0
x= -8/3 x=1
2007-05-29 21:33:10 · answer #3 · answered by leo 6 · 0⤊ 0⤋
3x^2 + 5x - 8 = 0
(3x + 8)(x - 1)
3x^2 +8x -3x -8
3x^2 + 5x - 8
x = 1 or x = -8/3
So the positive root is 1.
2007-05-29 21:36:53 · answer #4 · answered by Matthew H 2 · 0⤊ 1⤋
3x^2 + 5x - 8 = 0
x = -5 + - sq rt (25 - 4(3)(-8)) / 2(3)
x = -5 + - sq rt (25 + 96) / 6
x = -5 + - sq rt 121 / 6
x = -5 + - 11 /6
Pos root
x = 6/6 = 1
2007-05-29 21:33:00 · answer #5 · answered by richardwptljc 6 · 0⤊ 0⤋
3x² + 5x - 8 = 0
x = [- 5 ± â(25 + 96)] / 6
x = [- 5 ± â(121)]/ 6
x = [- 5 ± 11] / 6
x = 1 is +ve root.
2007-05-30 03:32:58 · answer #6 · answered by Como 7 · 0⤊ 0⤋