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Math Help - Radical Equations

  1. #1
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    Radical Equations

    The answer to the question below is x = 7. How does one get 7 in this sample question?

    sqrt{5x + 1} + 3x = 27
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  2. #2
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    Quote Originally Posted by magentarita View Post
    The answer to the question below is x = 7. How does one get 7 in this sample question?

    sqrt{5x + 1} + 3x = 27
    \sqrt{5x+1}+3x=27
    \sqrt{5x+1}=27-3x
    Square both sides.
    5x+1=9x^2-162x+729
    9x^2-167x+728=0
    a=9 b=-167 c=728
    Use the quadratic formula: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
    And you end up with  x=7 or  x=\frac{104}{9}

    One slight problem there is that 104/9 is completely wrong as you can tell by putting it in. However, 7 fits perfectly and is the only real solution as far as I can tell.
    Last edited by JeWiSh; March 8th 2009 at 03:41 PM.
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  3. #3
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    yes...

    Quote Originally Posted by JeWiSh View Post
    \sqrt{5x+1}+3x=27
    \sqrt{5x+1}=27-3x
    Square both sides.
    5x+1=9x^2-162x+729
    9x^2-167x+728=0
    a=9 b=-167 c=728
    Use the quadratic formula: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
    And you end up with  x=7 or  x=\frac{104}{9}

    One slight problem there is that 104/9 is completely wrong as you can tell by putting it in. However, 7 fits perfectly and is the only real solution as far as I can tell.
    I was going to use the quadratic formula but decided not to.
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