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Algebraic anomaly

  1. Apr 14, 2016 #1
    1. The problem statement, all variables and given/known data

    I want to find x that satisfies ##x-\sqrt{(x+2)^2} = 0##
    Using algebric method, I find x = -1
    But, x = -1 does not satisfy that equation...
    What's wrong with the calculation?

    2. Relevant equations
    (Basic algebra)

    3. The attempt at a solution

    ##x-\sqrt{(x+2)^2} = 0 \\
    x = sqrt{(x+2)^2} ##
    By squaring both sides , I\ get
    ##x^2 = (x+2)^2 \\
    x^2 = x^2 + 4x + 4 \\
    0 = 4x + 4 \\
    x = -1 \\
    ##

    What's wrong with this ?
     
    Last edited by a moderator: Apr 14, 2016
  2. jcsd
  3. Apr 14, 2016 #2

    blue_leaf77

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    The equation you want to solve does not have a solution to begin with.
     
  4. Apr 14, 2016 #3
    So, does it mean that it's necessary to plug the solution that I get from algebric calculation into the equation to confirm it's true??
     
  5. Apr 14, 2016 #4

    Mark44

    Staff: Mentor

    When you square both sides of an equation, there's the possibility that you'll get an equation whose solutions are not the same as the original equation. In this case, squaring introduced an extraneous solution.
     
  6. Apr 14, 2016 #5

    Mark44

    Staff: Mentor

    Yes
     
  7. Apr 14, 2016 #6
    Hmmm..
    So squaring both sides is a wrong attempt to find the solution ?
    I wonder if there are any article about to-dos and not-to-dos in solving particular algebraic equations...
     
  8. Apr 14, 2016 #7

    Mark44

    Staff: Mentor

    No, this is the right approach. It's just that when you apply any non-reversible operation (such as squaring, raising to fourth power, etc.), you need to check to see if you have introduced extraneous solutions.
    You might try doing a search for "extraneous solutions". Most algebra textbooks mention this in the section where equations involving radicals are presented.
     
  9. Apr 14, 2016 #8

    blue_leaf77

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    For this equation, you can actually still work in a way which does not introduce any extra solution. Note that you can cast the original equation into ##x-|x+2|=0##. Solve this equation for ##x<-2## and ##x>-2## to see if there is a solution in each region.
     
  10. Apr 14, 2016 #9
    Thanks a lot!

    x-x+2 = 0 (No solution)
    x+x+2=0
    2x+2=0
    x=-1 (Still an extraneous solution)
     
  11. Apr 14, 2016 #10

    blue_leaf77

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    x-x-2=0
    You forget that you are looking for a solution in ##x<-2##. The solution you find there lies outside this range.
     
  12. Apr 14, 2016 #11
    So it means no solutions, right ?
     
  13. Apr 14, 2016 #12

    blue_leaf77

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  14. Apr 15, 2016 #13

    SammyS

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    Suppose you were to solve ##\ x+\sqrt{(x+2)^2\,} = 0\ ## and you used the same method.

    That is equivalent to ##\ \sqrt{(x+2)^2\,} = -x\ .##

    Squaring both sides gives ##\ (x+2)^2 = x^2\ ##, the same as in your case, giving ##\ x=-1\ .## Here the solution is valid.
     
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