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Math Help - Find All Solutions to "x" In Radical Fraction

  1. #1
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    Find All Solutions to "x" In Radical Fraction

    Hi everyone!

    I've been struggling with this problem involving radical expression as part of my algebra assessment:

    [sqrt(x+1) + sqrt(x-1)]/[sqrt(x+1) - sqrt(x-1)] = 3 Solve for all possible solutions to "x".

    So far I've tried to simplify the expression by multiplying the denominator by its conjugate, like so:

    1) ([sqrt(x+1) + sqrt(x-1)]/[sqrt(x+1) - sqrt(x-1)]) * ([sqrt(x+1) + sqrt(x-1)]/[sqrt(x+1) + sqrt(x-1)]) = 3
    2) (x+1 + 2[sqrt(x+1) * sqrt(x-1)] + x-1)/[x+1 - (x-1)] = 3
    3) (2x + 2[sqrt(x+1) * sqrt(x-1)])/2 = 3
    4) (x + [sqrt(x+1) * sqrt(x-1)]) = 3
    5) ?


    This is where I'm stuck. I don't know how to proceed from here to solving for all solutions to "x", so if someone could help me from here I would appreciate it!
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  2. #2
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    Re: Find All Solutions to "x" In Radical Fraction

    $3=\dfrac {\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=$

    $\dfrac {\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}=$

    $\dfrac{(x+1)+2\sqrt{x^2-1}+(x-1)}{(x+1)-(x-1)}=$

    $\dfrac {2x+2\sqrt{x^2-1}}{2}=3$

    $x+\sqrt{x^2-1}=3$

    $\sqrt{x^2-1}=3-x$

    $x^2-1=9-6x+x^2$

    $6x=10$

    $x=\dfrac 5 3$

    This is the only real solution.
    Thanks from topsquark, Lexielai and HallsofIvy
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  3. #3
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    Re: Find All Solutions to "x" In Radical Fraction

    Wow it seems so obvious now! Thanks for the help
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