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Radical with negative radicand squared

  1. Aug 5, 2014 #1
    Can someone explain this for me.

    sqrt x^2 = x

    (sqrt x)^2 = x

    Yet sqrt -1^2 = |x| = -(-1) = 1

    (sqrt -1)^2 = (sqrt -1)(sqrt -1) = -1
     
  2. jcsd
  3. Aug 5, 2014 #2

    jedishrfu

    Staff: Mentor

    This topic gets discussed many times here in the past. The confusion centers on the sqrt(x^2) producing a positive root x and a negative root -x. The square root of a negative number is not defined if we deal strictly with Real numbers instead we must extend the definition to complex numbers where i^2 = -1. In that case, sqrt(-1) is either i or -i.

    This may help:

    http://mathbitsnotebook.com/Algebra1/Radicals/RADNegativeUnder.html
     
  4. Aug 5, 2014 #3
    I know about imaginary numbers but I don't think they apply here. Most texts I've looked at say the square root of a radicand squared is equal to the square of the radical. Yet, when you take the square root of negative number squared you get the positive root and when you square the square root of a negative number you get the negative number. I hope this makes sense, if I could write this out in symbolic form I'm sure it would be clearer.

     
  5. Aug 5, 2014 #4

    jedishrfu

    Staff: Mentor

    Given radicand x then you're saying: (sqrt(x))^2 = x

    so if x=4 then the statement would be either (sqrt(4))^2 = (2)^2 = 4 or (sqrt(4))^2 = (-2)^2 = 4 right?

    and if x = -4 then we get either:

    (sqrt(-4))^2 = (2i)^2 = -4 or (sqrt(-4))^2 = (-2i)^2 = -4

    so yes what you have said is true when complex numbers are used but when only real numbers are allows then
    the expression sqrt(x) where x<0 is undefined and so the stmt (sqrt(x))^2 = x is not valid.
     
  6. Aug 5, 2014 #5

    Mark44

    Staff: Mentor

    By convention, the square root of a positive number is positive, so ##\sqrt{x^2} = x## only when x ≥ 0. Most books show this identity: ##\sqrt{x^2} = |x|## to cover the case when x < 0.
    If they do, I believe you are missing some "fine print" -- namely the condition that x is nonnegative.
    Yes.
    Yes, but in the intermediate step (the square root of a negative number) you have an imaginary number.
     
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