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Math Help - Extra roots?

  1. #1
    Senior Member DivideBy0's Avatar
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    Extra roots?

    The problem is solve for x:

    \sqrt{x^2+1}+x^2+1=90

    If I do this in the 'normal' way, by isolating the square root and squaring both sides, I get:

    x^4-179x^2+7920=0

    (x^2-80)(x^2-99)=0

    x^2 = 80, 99

    This yields x = \pm 4\sqrt{5}, \pm 3\sqrt{11}

    Why did the two extraneous roots \pm 3\sqrt{11} appear?

    Thanks!

    (I realised you can immediately set y=\sqrt{x^2+1}, but that's not the point)
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  2. #2
    MHF Contributor red_dog's Avatar
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    Medgidia, Romania
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    5
    \sqrt{x^2+1}=89-x^2
    Before squaring both sides and solve for x, you must put the condition that the right-hand member is nonnegative.
    89-x^2\geq 0\Rightarrow x\in[-\sqrt{89},\sqrt{89}].
    So, the other two roots, \pm 3\sqrt{11} are not in this interval.


    Another way to solve the equation:
    Let \sqrt{x^2+1}=t, \ t>0.
    Then t^2+t-90=0\Rightarrow t_1=-10, \ t_2=9
    But t>0\Rightarrow t=9\Rightarrow x^2+1=81\Rightarrow x=\pm 4\sqrt{5}.
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  3. #3
    Senior Member DivideBy0's Avatar
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    Thanks, I guess that's something new I'll have to watch out for
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