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B PEMDAS & Roots

  1. Mar 5, 2016 #1
    I am having trouble understanding why the following expression is equal to 3:

    SQRT((-3)^2)

    What prevents me from cancelling the exponent 2 by the exponent 1/2 introduced by the square root, which yields a value of -3?

    Thanks for the help!
     
  2. jcsd
  3. Mar 5, 2016 #2

    Simon Bridge

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    What does SQRT() do?

    Usually the surd is defined so that ##\sqrt{x}## is the positive square root, when taken by itself like you seem to have.
    Computer scripts usually evaluate it the same way. When it appears in an expression, you can use the context to help.
    You can tie yourself in knots doing the other way.

    Put ##x=\sqrt{(-3)^2}## ... what values of x make this expression true?

    By order of operation you do the inner brackets first ... so you could write:
    ##x = ((-3)^2)^{1/2}## and still get the same out.
     
  4. Mar 6, 2016 #3

    Math_QED

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    SQRT(x^2) = |x|

    SQRT((-3)^2) = |-3| = 3

    I hope I could help.
     
  5. Mar 6, 2016 #4
    Okay... so from what I'm understanding so far you should work from the inside out when dealing with parentheses inside parentheses. This allows me to get the value of 3. However, suppose I define x = -3. I then have SQRT(x^2) which yields just x. This simplification yields -3 as the result of the expression. What gives?

    Thanks again for all the help.

    *EDIT: Reading some more online it looks as though a square root more explicitly defined has two solutions, both a positive and negative. This doesn't necessarily answer my question on correct order of operations in this scenario, however. Any help there is still appreciated.
     
  6. Mar 6, 2016 #5

    Simon Bridge

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    You did the maths wrong.
    x==-3, x^2 = 9, SQRT(9)=3 = -x
    i.e. if x<0, then SQRT(x^2)=-x
    i.e. SQRT(x^2) = |x|

    note: if x < 0, then ##\sqrt{x} = i\sqrt{-x}: i=\sqrt{-1}##
     
    Last edited: Mar 6, 2016
  7. Mar 7, 2016 #6

    Mark44

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    No (as Simon already said). ##\sqrt{(-3)^2} = \sqrt{9} = 3 \ne -3##
    No. The principal square root of a number x, which is denoted as ##\sqrt{x}##, is the positive number a such that ##a^2 = x##. I'm assuming here that x is a nonnegative real number, so that we don't go off into the imaginary numbers.
    Your question was already answered. ##((-3)^2)^{1/2}## should be evaluated from the inner-most parentheses outward.
     
  8. Mar 18, 2016 #7
    You can't cancel them because they aren't inverse functions over the domain of negative numbers. The square function isn't one to one. Both -3 and 3 map to the same number 9. So it doesn't have an inverse. It only has an inverse when taken over a limited domain (such as non-negative real numbers).
     
  9. Oct 10, 2016 #8
    You can simplify it under sqrt as,
    √(-3)2 ⇒ √(-3)×(-3) = √9 = ±3

    I hope it' ll help.
     
  10. Oct 10, 2016 #9

    Mark44

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    That last part is incorrect, but it is a very common mistake. √9 = 3. The symbol √9 represents the principal square root of 9, which is 3.
     
  11. Oct 11, 2016 #10

    Simon Bridge

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    ... for example: the quadratic formula has a ##\pm## before the root symbol [##2ax = -b^2\pm\sqrt{b^2-4ac}##]... which makes no sense if the ##\pm## is implicit in it.

    Now if you were to say ##x^2=9## then the solution is ##x=\pm\sqrt{9} = \pm 3## or ##x\in\{-3,3\}## ... that's where the ##\pm## gets used.
     
  12. Oct 11, 2016 #11
    ##\sqrt{(-3)^2}##
    ##=\sqrt{9}##
    =3
     
    Last edited by a moderator: Oct 12, 2016
  13. Oct 14, 2016 #12
    The problem is that √(x2) ≠ x, if we're considering all real numbers (-∞, ∞).

    The square root function is only defined over the non-negative real numbers, which is fine because the squaring function always gives us non-negative reals.

    The function f(x) = x2, takes us from any real number (-∞, ∞) to a non-negative real number [0, ∞).

    The function g(x) = √x, takes us from a non-negative real number [0, ∞) to a non-negative real number [0, ∞).

    Applying the square, followed by the square root, should give us a function that goes:

    g(f(x)) = ? : (-∞, ∞) -> [0, ∞) -> [0, ∞)

    However the expression 'x' as a function goes like:

    h(x) = x : (-∞, ∞) -> (-∞, ∞)

    Which is not what we expect from our composed function. We can resolve this by realizing that

    f(x) = x2 = |x|2

    In which case

    g(f(x)) = √(|x|2) = |x| : (-∞, ∞) -> [0, ∞)

    Take my explanation with a grain of salt, as I too am still learning the rules.
     
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