2007-07-05 11:48:21 · 6 answers · asked by Leo 1 in Science & Mathematics ➔ Mathematics
9x^2-9x=0
9x(x - 1) = 0
Therefore:
x = 0 or
x - 1 = 0, giving x = 1.
2007-07-05 11:52:44 · answer #1 · answered by Anonymous · 0⤊ 0⤋
First,find the multiplies:
9x^2-9x=0
9x(x-1) = 0
One part of the multiples has to be zero,therefore:
x-1 = 0 or 9x= 0
x= 1 or x = 0
2007-07-05 11:53:08 · answer #2 · answered by Leprechaun 6 · 0⤊ 0⤋
you have to solve it by factoring.
first, notice that the 9x repeats, so factor it out:
9x(x-1)=0
Now let 9x=0 and (x-1)=0.
Solve for x on both. So, you get x=0 and x=1
2007-07-05 11:54:30 · answer #3 · answered by EGD173 2 · 0⤊ 1⤋
The possible solutions are x = 0 or 1.
You solve it as follows. Factorize and rewrite the LHS as
9 x (x - 1) = 0.
Then either x = 0, or x - 1 = 0, that is x = 1.
Hence x = 0 or 1. QED
Live long and prosper.
2007-07-05 11:53:08 · answer #4 · answered by Dr Spock 6 · 0⤊ 0⤋
9x.(x - 1) = 0
x = 0 , x = 1
2007-07-09 10:49:14 · answer #5 · answered by Como 7 · 0⤊ 0⤋
9x(x-1)=0
x=0 and x=1
2007-07-05 11:54:50 · answer #6 · answered by nasser a 2 · 1⤊ 1⤋