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Math Help - Quadratic Equations

  1. #1
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    Quadratic Equations

    This question's troubling me...

    Solve this equation

    4^{1+x} + 4^{1-x} = 10


    Thanks in advance.
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  2. #2
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    \displaystyle 4^{1+x} + 4^{1-x} = 10

    \displaystyle 4\cdot 4^{x} + 4\cdot 4^{-x} = 10

    \displaystyle 4^x + 4^{-x} = \frac{5}{2}

    \displaystyle 4^x(4^x + 4^{-x}) = \frac{5}{2}(4^x)

    \displaystyle (4^{x})^2 + 1 = \frac{5}{2}(4^x)

    \displaystyle (4^x)^2 - \frac{5}{2}(4^x) + 1 = 0.


    Now let \displaystyle X = 4^x to get

    \displaystyle X^2 - \frac{5}{2}X + 1 = 0

    \displaystyle 2X^2 - 5X + 2 = 0

    \displaystyle 2X^2 - X - 4X + 2 = 0

    \displaystyle X(2X-1) - 2(2X-1) =0

    \displaystyle (2X-1)(X-2)=0

    \displaystyle 2X-1 = 0 or \displaystyle X-2=0

    \displaystyle X=\frac{1}{2} or \displaystyle X=2

    \displaystyle 4^x = \frac{1}{2} or \displaystyle 4^x = 2

    \displaystyle 2^{2x} = 2^{-1} or \displaystyle 2^{2x} = 2^1

    \displaystyle 2x = -1 or \displaystyle 2x = 1

    \displaystyle x = -\frac{1}{2} or \displaystyle x=\frac{1}{2}.
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  3. #3
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    Using hyperbolic cosine:

    4^{1+x} + 4^{1-x} = 10

    4(4^{x} + 4^{-x}) = 10

    4^{x} + 4^{-x} = \frac{5}{2}

    (e^{\ln(4)})^x + (e^{\ln(4)})^{-x} = \frac{5}{2}

    e^{x \ln(4)} + e^{-x \ln(4)} = \frac{5}{2}

    \displaystyle \frac{e^{x \ln(4)} + e^{-x \ln(4)}}{2} = \frac{5}{4}

    \cosh(x \ln(4)) = \frac{5}{4}

    x \ln(4) = \pm \cosh^{-1}(\frac{5}{4})

    x = \pm \cosh^{-1}(\frac{5}{4}) / \ln(4)

    x = \pm \frac{1}{2}
    Last edited by snowtea; December 16th 2010 at 08:39 PM.
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