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Extraneous roots

  1. Apr 3, 2008 #1
    Sorry for the length of the post, the problem I've included is not difficult but I wanted to have an example to help illustrate my question.
    solve:

    [tex]\sqrt{x}-\sqrt[4]{x} -2=0 [/tex]
    .
    .
    .
    [tex](x-16)(x-1)=0[/tex]

    The roots are 16 and 1, however when one puts them back into the original equation, 1 is found to be extraneous leaving 16 as the only solution. My question is, why do extraneous roots arise?
    I attempted to answer the question myself by reversing the above process and putting 1 in for x at each step to see when the equation becomes "invalid" for the extraneous root.

    [tex](x-16)(x-1)=0[/tex]

    [tex]x^{2}-17x+16=0[/tex]

    [tex]x^{2}-17x+16+25x=25x[/tex]

    [tex]x^{2}+8x+16=25x[/tex]

    [tex](x+4)^{2}=25x[/tex]

    [tex]x+4=5\sqrt{x}[/tex]

    [tex]x+4-4\sqrt{x}=5\sqrt{x}-4\sqrt{x}[/tex]

    [tex]x+4-4\sqrt{x}=\sqrt{x}[/tex]

    [tex](\sqrt{x}-2)^{2}=\sqrt{x}[/tex]

    [tex](\sqrt{x}-2)^{2}=\sqrt{x}[/tex] equation A

    [tex]\sqrt{x}-2=\sqrt[4]{x}[/tex] equation B

    [tex]\sqrt{x}-\sqrt[4]{x}-2=0[/tex]

    Putting 1 in for x in equation A works but B does not. It seems that going from A to B creates the problem. When one takes the square root of equation A the left side becomes

    [tex]((\sqrt{x}-2)^{2})^{\frac{1}{2}}[/tex]

    If I understood CompuChip's answer correctly to one of my previous posts, the inner to outer priority is not followed. If 1 is in for x, then -1 is the value in the first set of parenthesis and then -1 squared is 1, and then the square root is also 1. However if 1 is not in for x , since the roots are not known when one first goes through the problem, the squared to the 1/2 power gives what's in the parenthesis to the first power, which is just what's in the parenthesis. Then when 1 is in for x, we have -1 to the first power which is -1. The order of operations makes a difference for x=1 but does not for x=16.
    Is it true then, that extraneous roots arise because some mathematical operation is violated for that root?
    Thanks for any answers.
     
  2. jcsd
  3. Apr 3, 2008 #2
    because equation B is not your original equation, you changed the power, and because of that you changed the roots
     
  4. Apr 4, 2008 #3

    Gib Z

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    Homework Helper

    [tex]x = 5, x^2= 25, x=5, -5[/tex]. Exactly the same as that, but more disguised =] In this same one, when you squared it, you introduced the erroneous negative square root when only the positive root applies.
     
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