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Thread: Solve Equation

  1. #1
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    Solve Equation

    Solve:

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

    $\displaystyle x-\sqrt{x+4}=2$ becomes $\displaystyle x^2-x+4=4$ which becomes $\displaystyle x^2=x$ 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:

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

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

    $\displaystyle 0 = x^2 - 5x$

    $\displaystyle 0 = x(x - 5)$

    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.

    $\displaystyle x-\sqrt{x+4}=2$ becomes $\displaystyle x^2-x+4=4$ which becomes $\displaystyle x^2=x$ 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
    $\displaystyle (x - \sqrt{x + 4})^2 \neq x^2 - (\sqrt{x + 4})^2$

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

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

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

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