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Math Help - Problem on square root...

  1. #1
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    Problem on square root...

    Find the value of x in \sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}=x

    I want to know where else I can find such questions and some hints to solve such questions if anyone can provide it would be a lot of help.
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  2. #2
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    Hello siddscool19

    I don't have an algebraic solution to this problem. If you try squaring both sides to get rid of square roots, you'll end up with an equation involving x^{16}. So I think you'll probably have to resort to a bit of intelligent trial and error.

    If this equation is going to have any exact integer solutions then 3x will have to be a perfect square. The only candidates are 3x = 0, 9, 36, 81, ..., giving possible values of x as 0, 3, 12, 27, ...

    In fact x = 0 and x = 3 are solutions, and I'm certain they are the only ones.

    I cheated a bit and used a spreadsheet to have a look at the function
    f(x)= \sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}

    When 0<x<3, f(x)>x and for x > 3, f(x)<x and is increasing more slowly than x. So I'm sure that there are no further solutions.

    I don't know whether anyone else can come up with a more analytical solution.

    Grandad
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  3. #3
    MHF Contributor red_dog's Avatar
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    We have x\geq 0

    A solution is x=0

    Let x>0. Devide the equation by x:

    \displaystyle\sqrt{\frac{1}{x}+2\sqrt{\frac{1}{x^3  }+2\sqrt{\frac{1}{x^7}+2\sqrt{\frac{3}{x^{15}}}}}}  =1

    Let y=\frac{1}{x} and f(y)=\sqrt{y+2\sqrt{y^3+2\sqrt{y^7+2\sqrt{3y^{15}}  }}}-1

    f is strictly increasing and has a unique root: y=\frac{1}{3}\Rightarrow x=3

    So, the solutions of the equation are x=0, \ x=3
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