# Homework Help: Intersection of 2 curves

1. Dec 5, 2007

### fk378

I'm trying to find where x+y=1 meets x=2(y^2)

To solve for y I set up:
2(y^2)=1-y
2(y^2)+y=1
y(2y+1)=1
I have y=1 and 2y+1=1
for 2y+1=1, 2y=0 so y=0
y=0,1

But, I notice that my teacher did:
2(y^2)+y-1=0
(2y+1)(y-1)=0
y=-1, 1/2

Why are these 2 methods bringing about different answers? Shouldn't they be the same?

Last edited: Dec 5, 2007
2. Dec 5, 2007

### Dick

Neither y=0 nor y=1 solves y*(2y+1)=1. Try to think about that a little.

Last edited: Dec 5, 2007
3. Dec 5, 2007

### fk378

That is true...if I plug it in, it doesn't work....so how come if I solve for y, those are the 2 answers I get? And whats the difference between solving for it by setting it equal to 1, and setting it equal to 0?

4. Dec 5, 2007

### Dick

Because a*b=0 means a=0 or b=0. a*b=1 does not mean a=1 or b=1. After all, (1/2)*2=1 and neither of those is equal to 1.

Last edited: Dec 5, 2007
5. Dec 6, 2007

### HallsofIvy

If a and b are numbers and ab= 0, then either a= 0 or b= 0-the crucial point is that 0 divided by anything (other than 0) is 0. If a is not 0, divide both sides of the equation by a and you get b= 0. If b is not 0, divide both sides of the equation by b and you get a= 0.

That is not true for 1 or any other non-zero number. If ab= 1, then it might be that a= 1/2, b= 2. Or a= 1/4, b= 4, or ... many other possibilities. That is why, when factoring to solve an equation you MUST get it equal to 0.