# Homework Help: Simple Algebra

1. Mar 1, 2014

### emergentecon

1. The problem statement, all variables and given/known data

Solve for x

x(x-1)=0

2. Relevant equations

3. The attempt at a solution

x = 0 and x = 1

what I am trying to understand is the logic behind the x=0?
could someone please explain that to me?

2. Mar 1, 2014

### phinds

How much is zero times ANYTHING ?

3. Mar 1, 2014

Zero

4. Mar 1, 2014

### emergentecon

Are you dividing by x and x-1 to get the answer?

5. Mar 1, 2014

[offtopic]
This is $x^2-x=0$
hmm?
So $x^2=x$
How is this possible?
[/offtopic]

6. Mar 1, 2014

### emergentecon

Well if x = 0 and x = 1 then it holds?

7. Mar 1, 2014

$(x)(x-1)=0$
As you do with Quadratic equations,
$x=0$
$x-1=0$
so $x=1$.

That's it.Right?phinds

Last edited: Mar 1, 2014
8. Mar 1, 2014

### ChrisVer

you have a number on the left, and a number on the right...
x(x-1) is a number, so is 0... and you want them to be equal...
when can x(x-1) be equal to 0?
you have two possibilities....
either x=0, so you will have 0*(0-1)=0*(-1)=0
or x=1, so you will have 1*(1-1)=1*0=0
so in both these cases you achieved what the equation asked for you x(x-1)=0
you didn't divide,multiply or anything....

9. Mar 1, 2014

You don't have to guess things.That's done in the way I mentioned above.

10. Mar 1, 2014

### ChrisVer

it's not guessing... it's in fact what happens with factorizing anything (eg a polynomial equation)

11. Mar 1, 2014

### HallsofIvy

The obvious fact that if x= 0 then x(x- 1)= 0 and that if x= 1 then x-1= 0 so x(x- 1)= 0 shows that x= 0 and x= 1 are solutions but does NOT show that they are the only solutions. For example in the "integers modulo 6" it is true that 0 times anything is 0 so that x= 0 is a solution to 3x= 0 but so is x= 2.

For that you need the "zero factor property" some times phrased as "the set of real numbers (or complex numbers) does not have "zero divisors":

If ab= 0 then either a= 0 or b= 0

Which is not true for the "integers modulo 6".

12. Mar 1, 2014

### Staff: Mentor

It seems like you're going backwards here, going from x2 - x = 0 to x2 = x.
The OP already had the left side of the equation in factored form (i.e., x(x - 1) = 0). Expanding the left side and adding x to both sides doesn't buy you anything. The important principle here is that if the product of two numbers is zero, then one or the other of the numbers has to be zero.

13. Mar 1, 2014