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Homework Help: Solving an equation

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data

    Solve [tex]x-\sqrt{x}-6=0[/tex]

    2. Relevant equations

    3. The attempt at a solution

    This can be factorised into [tex](\sqrt{x}-3)(\sqrt{x}+2)=0[/tex]

    so [tex]\sqrt{x}=3[/tex] , x=9 .

    [tex]\sqrt{x}=-2[/tex] ... Why is this root not acceptable ?

    I recalled that if

    [tex]x^2=4[/tex] , then [tex]x=\pm 2[/tex]

    but if [tex]x=\sqrt{4}[/tex] , then x=2

    Is this true ?
  2. jcsd
  3. Apr 26, 2010 #2


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    Homework Helper

    Yes, it is conventional that the square root is only the positive, so while [itex]x^2=4[/itex] and therefore [itex]x=\pm 2[/itex] ... if [itex]\sqrt{x}=-2[/itex] then this cannot be solved with x=4 since [itex]\sqrt{4}=2[/itex] and not [itex]\sqrt{4}=\pm 2[/itex].
  4. Apr 26, 2010 #3
    thanks , but why is it so , since when you square x=-2 , you still get 4 ?
  5. Apr 26, 2010 #4


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    Because (-2)(-2)= (-1)(-1)(4) and (-1)(-1)= +1.
  6. Apr 26, 2010 #5
    thanks i know that , i just wonder why is it conventional for the square root positive only , why isn't the negative taken into consideration in this case ?
  7. Apr 26, 2010 #6
    The square root is a function. Do you know the definition of a function?
  8. Apr 27, 2010 #7
    I think so , say [tex]y=\sqrt{x}[/tex] , and any input would generate only one image , why cant this image be -2 instead of 2 if the input is 4
  9. Apr 27, 2010 #8


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    Sorry, I misunderstood your question. A "function" can give only one value for each value of x so we must choose either the positive or negative root0. We could define [itex]\sqrt{x}[/itex] to be the negative root but then we would have problems with "compositions" such as [itex]\sqrt{\sqrt{x}}[/itex] since the square root of a negative number is not defined in the real number system.
  10. Apr 27, 2010 #9
    thanks a lot Hallsofivy , i finally understood.
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