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Intersection of 2 curves

  1. Dec 5, 2007 #1
    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. jcsd
  3. Dec 5, 2007 #2

    Dick

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    Neither y=0 nor y=1 solves y*(2y+1)=1. Try to think about that a little.
     
    Last edited: Dec 5, 2007
  4. Dec 5, 2007 #3
    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?
     
  5. Dec 5, 2007 #4

    Dick

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    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
  6. Dec 6, 2007 #5

    HallsofIvy

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    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.
     
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