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Math Help - Solve Equation

  1. #1
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    Solve Equation

    Hii !

    Can You Help Me For This Equation ,

    Solve in IR the Equation :

    x=a-\sqrt{a-\sqrt{x}}

    following the values of parameter a ( a \in \mathbb{R^+} )
    Last edited by Perelman; February 1st 2010 at 11:50 AM.
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  2. #2
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    Well

    What have you tried so far? Have you subtracting an a from both sides, and then square both sides?
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  3. #3
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    x= a-\sqrt{a-\sqrt{x}}
    \Longleftrightarrow a-x=\sqrt{a-\sqrt{x}}
    \Longleftrightarrow a^2-2ax+x^2=a-\sqrt{x}

    after ?
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  4. #4
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    Quote Originally Posted by Perelman View Post
    x= a-\sqrt{a-\sqrt{x}}
    \Longleftrightarrow a-x=\sqrt{a-\sqrt{x}}
    \Longleftrightarrow a^2-2ax+x^2=a-\sqrt{x}

    after ?
    Well if you want to solve for x or a? If it is x then subtract another a from each side and square them again (a^2-2ax+x^2-a)(a^2-2a+x^2-a) = x
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  5. #5
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    Quote Originally Posted by Henryt999 View Post
    Well if you want to solve for x or a? If it is x then subtract another a from each side and square them again (a^2-2ax+x^2-a)(a^2-2a+x^2-a) = x
    Solve the Equation, by following the values of parameter a ( a \in \mathbb{R^+} )
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  6. #6
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    Salut !

    x=a-\sqrt{a-\sqrt{x}}

    \sqrt{x}=\sqrt{a-\sqrt{a-\sqrt{x}}}

    a-\sqrt{x}=a-\sqrt{a-\sqrt{a-\sqrt{x}}}

    \sqrt{a-\sqrt{x}}=\sqrt{a-\sqrt{a-\sqrt{a-\sqrt{x}}}}

    a-\sqrt{a-\sqrt{x}}=a-\sqrt{a-\sqrt{a-\sqrt{a-\sqrt{x}}}}

    x=a-\sqrt{a-\sqrt{a-\sqrt{a-\sqrt{x}}}}

    and so on ...

    Therefore taking the series defined by u_0 = a and u_{n+1}=f(u_n) where f(x) = a-\sqrt{x}

    x can be seen as the limit (if it exists) of \left(u_n\right)_n

    If it converges, it is towards l such as l = f(l) which can be solved by a quadratic

    At the end x = \frac{2a+1\pm \sqrt{4a+1}}{2} ... if I am not mistaken

    It seems to be "minus" but I let you see why !!

    All what I have written is to be checked ...
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