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Quadratic Equation

  1. Apr 24, 2012 #1
    solve √(5x-1) - √x=1 eq. 1
    transpose one of the radicals: √(5x-1)=√x+1
    Square: 5x-1=x+2√x+1
    Collect terms: 4x-2=2√x or 2x-1=√x
    Square: 4x^2-4x+1=x, x=1/4 , 1.

    x=1/4 is not the root by substitution to equation 1.

    My question what makes x =1/4 is not the root, only x=1 is the root.
  2. jcsd
  3. Apr 24, 2012 #2


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    The fact that [itex]\sqrt{5(1/4)- 1}= \sqrt{5/4- 1}= \sqrt{1/4}= 1/2[/itex]
    while [itex]\sqrt{1/4}= 1/2[/itex] also so that if x= 1/4 [itex]\sqrt{5x-1}- \sqrt{x}= 0[/itex], not 1.

    That is, it is not a solution because it does not satisfy the equation!

    Any time we square both sides of an equation, or multiply both sides of an equation by something involving "x", we may introduce "new" solutions that satisfy the new equation but not the original one.
  4. Apr 24, 2012 #3
    Does this occurs only for equations with radicals since normally we do not test the values.
  5. Apr 24, 2012 #4

    This may occur anytime we're trying to solve equations by methods that involve some operation that can

    give two or more different results. In this case, when you square stuff, the difference between positive and

    negative terms disappears and thus it is necessary, at the end of the process, to check each and every

    supposed solution in the original equation.

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