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  1. #1
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    solve

    solve the equation


    x\sqrt {1 + \frac {1}{x}} + \frac {1}{x}\sqrt {x + 2} = 1 + \sqrt {x + \frac {2}{x} + 3}
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  2. #2
    Moo
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    Hello,

    <br />
x\sqrt {1 + \frac {1}{x}} + \frac {1}{x}\sqrt {x + 2} = 1 + \sqrt {x + \frac {2}{x} + 3}<br />

    You can notice that \left(1+\frac 1x\right) \cdot (x+2)=\left(x+\frac 2x+3\right)


    1. Good (?) method
    \text{equation}<br />
\quad \Longleftrightarrow \quad \left(x \sqrt{1+\frac 1x}-1\right)-\left(\sqrt{\left(1+\frac 1x\right)(x+2)}-\frac 1x \sqrt{x+2}\right)=0

    x \left(\sqrt{1+\frac 1x}-\frac 1x\right)-\sqrt{x+2} \left(\sqrt{1+\frac 1x}-\frac 1x\right)=0

    (x-\sqrt{x+2})\left(\sqrt{1+\frac 1x}-\frac 1x\right)=0

    Therefore :

    \text{either } \left\{\begin{array}{ll} x-\sqrt{x+2}=0 \quad (1) \\ \sqrt{1+\frac 1x}-\frac 1x=0 \quad (2) \end{array} \right.

    (1) \implies x=\sqrt{x+2} \implies x^2-x-2=0 \implies (x+1)(x-2)=0
    \boxed{x=-1} or \boxed{x=2}

    (2) \implies \frac 1x=\sqrt{1+\frac 1x} \implies \frac 1{x^2}=1+\frac 1x \implies x^2+x-1=0
    \boxed{x=\frac{-1 \pm \sqrt{5}}{2}}


    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    2. Rough method
    Square both sides :

    x^2 \left(1+\frac 1x\right)+\frac{x+2}{x^2}+{\color{red}2 \sqrt{x+\frac 2x+3}}=1+\left(x+\frac 2x+3\right)+{\color{red}2 \sqrt{x+\frac 2x+3}}

    (x^2+x)+\left(\frac 1x+\frac 2{x^2}\right)=1+x+\frac 2x+3

    x^2-\frac 1x+\frac 2{x^2}-4=0

    Multiply by x^2 :

    x^4-4x^2-x+2=0

    Note that -1 is a solution, by looking at the coefficients.

    (x+1)(x^3-x^2-3x+2)=0

    Here, you can use the rational roots theorem... or see that 2 is a solution.

    (x+1)(x-2)(x^2+x-1)=0
    ...
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Moo View Post
    (1) \implies x=\sqrt{x+2} \implies x^2-x-2=0 \implies (x+1)(x-2)=0
    \boxed{x=-1} or \boxed{x=2}

    (2) \implies \frac 1x=\sqrt{1+\frac 1x} \implies \frac 1{x^2}=1+\frac 1x \implies x^2+x-1=0
    \boxed{x=\frac{-1 \pm \sqrt{5}}{2}}
    In both cases x can't be negative since we have x=\sqrt{x+2}\geq0 or \frac 1x=\sqrt{1+\frac 1x}>0.
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