(x^3-16x)
x(x^2-16x) <=== At this step, there shouldn't be an x with the 16.
x(x-4)(x+4)
You factored both and right! So what are the solutions for ?
For the other one, remember it's under the operator, so factoring won't do anything for now.
You need to solve so remember that and so set .
Sorry, I'm not following here. Could you explain what you're solving? If it's solving , then does not imply that either or . Also, see the note.
You were solving x(x-4)(x+4) = 0 and you got x = 4 or x = -4 as solutions. What's the other solution?
You're right, of course. I'm not sure whether the OP saw that x = 0 is not in the domain (I didn't for sure).
I was thinking of finding all the solutions for the x²-1 = 1 and x³-16x = 0 then discarding the unfit ones.
Last edited by TheCoffeeMachine; Mar 6th 2011 at 09:00 AM.