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Math Help - Problem #19: Exponent Solution

  1. #1
    Forum Admin topsquark's Avatar
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    Problem #19: Exponent Solution

    Here's a more straightforward problem for College Algebra level.

    Let a > 1 be an integer. Find all integers x, such that

     \left ( \sqrt{ a + \sqrt{a^2 - 1} } \right ) ^x + \left ( \sqrt{ a - \sqrt{a^2 - 1} } \right ) ^x = 2a

    (Source: Singapore Mathematics Olympiad)

    -Dan
    Last edited by topsquark; August 17th 2014 at 02:59 PM. Reason: Need a course on basic spelling!
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    Forum Admin topsquark's Avatar
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    Re: Problem #19: Exponent Solution

    Hint:
    Spoiler:

    Let y = \sqrt{a + \sqrt{a^2 - 1}}. What is 1/y?


    -Dan
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    Forum Admin topsquark's Avatar
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    Re: Problem #19: Exponent Solution

    No takers this time? Here's my solution:

    Spoiler:

     \left ( \sqrt{ a + \sqrt{a^2 - 1} } \right ) ^x + \left ( \sqrt{ a - \sqrt{a^2 - 1} } \right ) ^x = 2a

    Let y = \sqrt{ a + \sqrt{a^2 - 1} }

    Then
    \frac{1}{y} = \frac{1}{\sqrt{ a + \sqrt{a^2 - 1} }} = \frac{1}{\sqrt{ a + \sqrt{a^2 - 1} }} \cdot \frac{\sqrt{ a - \sqrt{a^2 - 1} }}{ \sqrt{ a - \sqrt{a^2 - 1} }    }

    \frac{1}{y} = \frac{ \sqrt{ a - \sqrt{a^2 - 1} }}{ \sqrt{a^2 - (a^2 - 1)}} = \sqrt{ a - \sqrt{a^2 - 1} }

    Putting this into the original equation gives:
    y^x + y^{-x} = 2 a

    Multiplying both sides by y^x gives:

    y^{2x} + 1 = 2a y^x

    or
    y^{2x} - 2a y^x + 1 = 0

    This is a quadratic equation in y^x so we may apply the quadratic formula:

    y^x = \frac{(2a) \pm \sqrt{(2a)^2 - (4)(1)(1)}}{(2 \cdot 1)}

    After some simplification we get
    y^x = a \pm \sqrt{a^2 - 1}

    So we have two cases:
    y^x = a + \sqrt{a^2 - 1} = y^2, which leads to x = 2,

    or
    y^x = a - \sqrt{a^2 - 1} = y^{-2}, which leads to x = -2.

    You can easily back-substitute and you will find that both solutions fit the original equation, so x = \pm 2.


    -Dan
    Last edited by topsquark; August 24th 2014 at 03:45 PM.
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    Re: Problem #19: Exponent Solution

    But the answer should be \pm 2

    y^x+y^{-x}=y^2+y^{-2}

    y^x-y^2+y^{-x}-y^{-2}=0

    \left(y^x-y^2\right)\left(1-y^{-x}y^{-2}\right)=0

    which leads to x=\pm 2
    Last edited by Idea; August 24th 2014 at 08:10 AM.
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    Forum Admin topsquark's Avatar
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    Re: Problem #19: Exponent Solution

    Quote Originally Posted by Idea View Post
    But the answer should be \pm 2

    y^x+y^{-x}=y^2+y^{-2}

    y^x-y^2+y^{-x}-y^{-2}=0

    \left(y^x-y^2\right)\left(1-y^{-x}y^{-2}\right)=0

    which leads to x=\pm 2
    I have fixed the problem in my derivation. Thanks for the catch!

    -Dan
    Last edited by topsquark; August 24th 2014 at 03:46 PM.
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