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Thread: Solve the radical equation

  1. #1
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    Solve the radical equation

    Hi all,

    I need help/directions on how to solve this problem:



    Thank you much.
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  2. #2
    Super Member Quacky's Avatar
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    Just to clarify, do you mean $\displaystyle \sqrt3x-2$ or $\displaystyle \sqrt{3x}-2$ or $\displaystyle \sqrt{3x-2}$?
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  3. #3
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    Oh, sorry, the third one.

    The square root of 3x-2
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  4. #4
    Super Member Quacky's Avatar
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    Ok. Here's my attempt:

    $\displaystyle x-\sqrt{3x-2}=4$
    $\displaystyle x-4=\sqrt{3x-2}$
    $\displaystyle (x-4)^2=\sqrt{3x-2}^2$
    $\displaystyle x^2-8x+16=3x-2$
    $\displaystyle x^2-11x+18=0$
    $\displaystyle (x-9)(x-2)=0$
    $\displaystyle x=2 $ or $\displaystyle x=9$

    x=2 is a solution for the negative result of , because when I square root a number, the result could be positive or negative. $\displaystyle (-2)^2=2^2$ so square rooting also gives two possible solutions. The reason I bring this up is that otherwise, when checking it does not seem to work.
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  5. #5
    Super Member Quacky's Avatar
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    Sorry about the latex errors, I've corrected them now.
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