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Solving an equation

  1. Oct 2, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]x = \sqrt{4 - 3x}[/tex]

    3. The attempt at a solution
    [tex]x^2 = 4 - 3x[/tex]
    [tex]x^2 + 3x - 4 = 0[/tex]
    [tex](x+4)(x-1) = 0[/tex]

    [tex]x + 4 = 0[/tex]
    [tex]x = -4[/tex]

    [tex]x -1 = 0[/tex]
    [tex]x = 1[/tex]

    Plugging back into the original equation, -4 doesn't work and 1 does work. But, if I plug it into the quadratic they both work. So which is it?
     
  2. jcsd
  3. Oct 2, 2008 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi Sheneron! :smile:

    (have a square-root: √ and a square: ² :smile:)

    -4 would work if -4 = √(16).

    But of course √ is defined to be ≥ 0.

    So you "lose" any negative solutions. :smile:

    (compare, for example, x = √1 and x² = 1 … they look the same, but the only solution to the first is x = 1, while the solution to the second is x = ±1 :wink:)
     
  4. Oct 2, 2008 #3

    statdad

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    Homework Helper

    You must check the "solutions" you obtain in the original equation, not in one that comes from squaring that original equation. The reason is this: your two numbers come from a statement is

    [tex]
    a^2 = b^2
    [/tex]

    From this statement alone it is not possible to claim that

    [tex]
    a=b
    [/tex]

    automatically follows. Your result of [tex] x = -4 [/tex] is an illustration of this: clearly

    [tex]
    -4 \ne 4 = \sqrt{4 - 3(-4)}
    [/tex]

    but

    [tex]
    16 = (-4)^2 = \left(\sqrt{4-3(-4)}\right)^2
    [/tex]
     
  5. Oct 2, 2008 #4
    thank ye
     
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