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Squaring both sides of an equation- extraneous solutions

  1. Nov 11, 2017 #1

    DS2C

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    1. The problem statement, all variables and given/known data
    Looking for an explanation as to why, when we square both sides of an equation, we can get extraneous solutions. That is, why can we square both sides of an equation, and sometimes the solutions we get are not true.

    In my book, it gets a little wordy and doesn't make a lot of sense. It says that "If both sides of an equation are raised to the same power, all solutions of the original equation are among the solutions of the new equation. This does not say that raising both sides of an equation to a power yields an equivalent solution."
    This makes no sense.

    2. Relevant equations
    An example:
    $$\sqrt {4 - x} = x - 2$$
    $$\left(\sqrt{4 - x}\right)^2 = \left(x - 2\right)^2$$
    $$x = 0 ~or~ x = 3$$




    3. The attempt at a solution
    0 Results in a false statement where checked, and 3 results in a true statement.

    Solution is {3}

    But why does 0 result as an extraneous solution? We squared both sides of the equation, leaving it balanced.
     
  2. jcsd
  3. Nov 11, 2017 #2

    haruspex

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    Raising to an even power loses information. Both x and -x turn into x2. x=1 has one solution, x2=1 has two.
     
  4. Nov 11, 2017 #3

    SammyS

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    So, you might ask, "Why do we have this solution, ##\ x=0\,,\ ## which turns out to be extraneous?"

    Plugging in ##\ x=0\ ## gives ##\ \sqrt{4}\ ## on the left hand side. There are two numbers which give 4 upon being squared, 2 and −2. If you pick the −2, then you match the right hand side. However, the radical symbol indicates that we only want the principal square root and that's what makes this solution be extraneous

    Notice that squaring either of the following equations gives the same result.
    ## \sqrt {4 - x} = x - 2 ##

    ## -\sqrt {4 - x} = x - 2 ##​

    ##\ x=0\ ## is the solution to the second of these equations..
     
  5. Nov 11, 2017 #4

    DS2C

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    Let me see if I understand-
    If we take ##2^2##, and ##\left(-2\right)^2##, we get 4 in both cases. However we don't know if 2 or -2 was used to get this 4 as it could have been either. Is this correct?
     
  6. Nov 11, 2017 #5

    haruspex

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    Yes.
     
  7. Nov 11, 2017 #6

    DS2C

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    Ok thanks for the help guys. Very well explained.
     
  8. Nov 11, 2017 #7

    Mark44

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    If the operation you apply to both sides of the equation are reversible, then you won't get extaneous solutions. So adding the same quantity to both sides, subtracting the same quantify from both sides, multiplying both sides by the same nonzero number are examples of these kinds of operations. These are all reversible operations -- operations that are one-to-one, meaning that if the operations are represented as functions, these functions have inverses.

    Squaring both sides is not an invertible operation, for the reasons already given. haruspex mentioned raising both sides to an even power as an operation that isn't invertible. If you raise both sides to an odd power, such as by cubing both sides, that's an operation that is invertible, provided we limit the discussion to the real numbers. For example, the equations ##x = 2## and ##x^3 = 8## are equivalent, for real numbers x. (If we also allow complex numbers as solutions, then the second equation has two more solutions than the first.)
     
  9. Nov 12, 2017 #8

    DS2C

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    Move to my university and teach my math classes.
     
  10. Nov 12, 2017 #9

    Mark44

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    I'll be down your way next week, but just for a visit. I'm already teaching part-time at a CC in Washington state.
     
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