x = 3 is incorrect, or at least, not fully correct.
x(x-3) is quite obviously a second degree polynomial as (x)(x)= x^2
you would expect 2 solutions...
as u know, if ab = 0, the either a or b or both = 0, think about it, 2 nonzero numbers cant give 0 when multiplied.
so either x-3 = 0 in which case x = 3
or x = 0 in which case obviously x = 0
check the answers:
0(0-3) = 0 so x = 0 is ok
3(3-3) = 3(0) = 0 so its ok too
dividing by zero in this type of problem is very dangerous unless is explicitly stated that x is not 0.
hope this helps.
and dont answer to questions unless u know what ur doing...
