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Math Help - Solve Equation

  1. #1
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    Solve Equation

    Solve:

    x-{\sqrt{x+4}}=2
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  2. #2
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    This equation is somewhat ambiguous.

    x-\sqrt{x+4}=2 becomes x^2-x+4=4 which becomes x^2=x so therefore, x must be equal to 1 or 0

    that's the way I would have done it. This isn't certain though. Can we get clarification?
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by acevipa View Post
    Solve:

    x-{\sqrt{x+4}}=2
    Isolate the square root:
    -\sqrt{x + 4} = -x + 2

    Now square both sides:
    x + 4 = (-x + 2)^2 = x^2 - 4x + 4

    0 = x^2 - 5x

    0 = x(x - 5)

    So x = 0 or x = 5. Always check these in your original equation. I get that only the x = 5 is a solution.

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mrbuttersworth View Post
    This equation is somewhat ambiguous.

    x-\sqrt{x+4}=2 becomes x^2-x+4=4 which becomes x^2=x so therefore, x must be equal to 1 or 0

    that's the way I would have done it. This isn't certain though. Can we get clarification?
    Please note that
    (x - \sqrt{x + 4})^2 \neq x^2 - (\sqrt{x + 4})^2

    Make sure you FOIL these out so that you can see how they square. The general rule is
    (a \pm b)^2 = a^2 \pm 2ab + b^2

    -Dan
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  5. #5
    Math Engineering Student
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    Quote Originally Posted by acevipa View Post
    Solve:

    x-{\sqrt{x+4}}=2
    Before solving equations like this one, we require certain conditions:
    • x\ge-4.
    • x\ge2.

    Hence, we require that x\ge2. Now, by makin' the straightforward calculations we get x^2-5x=0, which is Dan's result. From here x=0 or x=5, but the answer must lie on [2,\infty) (our requirement), so x=5.
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