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B Solving equations with radicals (extraneous solutions)

  1. Aug 1, 2016 #1
    I am solving the equation ##\sqrt{x + 3} + 4 = \sqrt{8x + 1}##. I understand that , generally, to solve it, we have to eliminate the radicals by isolating a radical expression to one side and then squaring both sides of the equation.

    I end up obtaining two solutions: ##x = 6## and ##x = 22/49##. Plugging these into the original equation, I find that ##x = 6## works, but ##x = 22/49## does not. I understand that the latter is termed an extraneous solution. My question is, how do these extraneous solutions arise? Does it have something to do with the fact that we lose information about the original equation when we square both sides?
     
  2. jcsd
  3. Aug 1, 2016 #2

    mfb

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    2016 Award

    Staff: Mentor

    Right, you lose the sign.

    x = 5 has just one solution, but x2 = 25 has two. Your example is a more complex version of the same effect.
     
  4. Aug 1, 2016 #3

    disregardthat

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    Squaring both sides of an equation is not a reversible operation. Simply put, if ##a = b##, then ##a^2 = b^2##. But if you only know that ##a^2 = b^2##, then you cannot conclude that ##a = b##. Here there are two possibilities: ##a = b## and ##a = -b##. You have to be aware of what operations are reversible, and what operations are not. If all your operations are reversible, you don't have to check your solution. If some of them are not (such as squaring), then you have to check that.
     
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