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Solving an irrational equation

  1. Mar 31, 2015 #1
    I have the irrational equation ##\sqrt{x - 1} + \sqrt{2 - x} = 0##, which has no real solutions. However, when I try to solve the equation, I get a real solution, that is:

    ##\sqrt{x - 1} + \sqrt{2 - x} = 0##
    ##\sqrt{x - 1} = -\sqrt{2 - x}##
    ##(\sqrt{x - 1})^{2} = (-\sqrt{2 - x})^{2}##
    ##(\sqrt{x - 1})(\sqrt{x - 1}) = (-\sqrt{2 - x})(-\sqrt{2 - x})##
    ##x - 1 = 2 - x##
    ##x = \frac{3}{2}##

    What am I doing wrong here? In which step am I making a mistake?
     
  2. jcsd
  3. Mar 31, 2015 #2

    Mark44

    Staff: Mentor

    When you square both sides of an equation, there's the possibility of getting an extraneous solution, one that isn't a solution of the original equation. Here's a simple example.

    x = -2 -- solution set is {-2}
    Square both sides:
    x2 = (-2)2 = 4 -- solution set is now {-2, 2}

    By squaring both sides, the solution set went from {-2} to {-2, 2}. As far as the original equation is concerned, the solution x = 2 is extraneous.

    The bottom line is: Whenever you square both sides of an equation, you need to check that any solution you get satisifies the original equation.
     
  4. Mar 31, 2015 #3

    symbolipoint

    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    Checking your found value in the original equation, decide if that value works or not.
     
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