How is this solved??
x^2-3x>=0
2007-07-17 19:44:48 · 5 answers · asked by Joey 1 in Science & Mathematics ➔ Mathematics
Here's one method to solve it.
|y| >= c implies y >= c OR y <= -c.
x^2 - 3x >= 0
Complete the square.
x^2 - 3x + 9/4 >= 9/4
(x - (3/2))^2 >= 9/4
| x - (3/2) | >= 3/2
x - (3/2) >= 3/2 OR x - (3/2) <= -3/2
x >= 3 OR x <= 0
2007-07-17 19:52:22 · answer #1 · answered by Puggy 7 · 0⤊ 0⤋
x^2-3x>=0
x( x - 3) > = 0
x > = 0 and x >= 3 are the solutions.
2007-07-18 02:56:29 · answer #2 · answered by Swamy 7 · 0⤊ 0⤋
x^2-3x>=0
factorizing
x(x-3)>=0
therefore
x<=0;x>=3
2007-07-18 02:55:23 · answer #3 · answered by niki einstien 2 · 0⤊ 0⤋
x^2-3x>=0
x(x-3)>=0
x >=0
and x-3>=0 means x >3
2007-07-18 02:52:44 · answer #4 · answered by Jain 4 · 0⤊ 0⤋
x^2>=3x
x^2/x>=3
x>=3
It seems my answer is different to others although I still think I'm right.
2007-07-18 03:48:17 · answer #5 · answered by A.P 1 · 0⤊ 0⤋