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Math Help - Solving for x

  1. #1
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    Solving for x

    \sqrt{1+\sqrt{x-1}} + \sqrt{4-\sqrt{x-1}} = 3

    A = \sqrt{x-1}

    1 + A + 2\sqrt{(1+A)(4-A)} + 4 - A = 3

    2\sqrt{(1+A)(4-A)} = -2

    \sqrt{(1+A)(4-A)} = -1

    (1+A)(4-A) = 1

    4 +3A - A^{2} = 1

    3A - A^{2} = -3

    A^{2} -3A = 3

    x - 1 -3\sqrt{x-1} = 3

    x -3\sqrt{x-1} = 4

    -3\sqrt{x-1} = 4 - x

    9(x-1) = x^{2} - 8x + 16

    x^2 - 17x + 25 = 0

    And now I'm stuck. What did I do wrong?
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  2. #2
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    From A^2 - 3A = 3

    A^2 - 3A + \left(-\frac{3}{2}\right)^2 = 3 + \left(-\frac{3}{2}\right)^2

    \left(A - \frac{3}{2}\right)^2 = 3 + \frac{9}{4}

    \left(A - \frac{3}{2}\right)^2 = \frac{21}{4}

    A - \frac{3}{2} = \pm \frac{\sqrt{21}}{2}

    A = \frac{3}{2} \pm \frac{\sqrt{21}}{2}.


    Therefore \sqrt{x - 1} = \frac{3}{2} \pm \frac{\sqrt{21}}{2}

    x - 1 = \left(\frac{3}{2} \pm \frac{\sqrt{21}}{2}\right)^2

    x - 1 = \frac{9}{4} \pm \frac{3\sqrt{21}}{2} + \frac{21}{4}

    x - 1 = \frac{15}{2} \pm \frac{3\sqrt{21}}{2}

    x = \frac{17}{2} \pm \frac{3\sqrt{21}}{2}.


    Therefore the two solutions are

    x = \frac{17 - 3\sqrt{21}}{2} and x = \frac{17 + 3\sqrt{21}}{2}.
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  3. #3
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    Sorry for the hassle. I just discovered that I forgot to do 3^2 in the right sight of the second equation.

    The answer is
    (x-10)(x-1) = 0
    x = 10, 1
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