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Homework Help: Square root problem

  1. Sep 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Can someone explain to me how [tex]|-4|= \sqrt{(-4)^{2}}[/tex]

    I'm wondering why you can't cancel out the square root sign and the square above the -4, to leave you with -4.

    3. The attempt at a solution

    I know this has something to do with the absolute value of -4, being 4, but I just don't completely understand why you can't cancel out the square root sign and the square.
  2. jcsd
  3. Sep 8, 2009 #2
    |-4| ≠ √(-4²) (because |-4| = 4 and √(-4²) = √(-1(4²)) = √(-1(16)) = √(-16) = 4i, where i = √(-1) is the imaginary number). Therefore, trying to explain how |-4| = √(-4²) is meaningless.
  4. Sep 8, 2009 #3
    It's not meaningless. In R, a square root cannot be a negative number, hence the absolute value sign.
  5. Sep 8, 2009 #4
    Because |-4|=|[itex]\pm[/itex]4|=4. Why it is like that?

    Because √(-4)2=√16=4 but also √42=√16=4.

    You need to consider both positive and negative value for 4.
    Last edited: Sep 8, 2009
  6. Sep 8, 2009 #5
    kbaumen, I suspect that the original poster had edited his/her post. Before he/she edited his/her post, it was √(-4²), and go to this http://hotmath.com/images/gt/lessons/genericalg1/parabola_width.gif" [Broken], and it's obvious that y > 0 for any x values that is x ≠ 0), and |-4| = 4.
    Last edited by a moderator: May 4, 2017
  7. Sep 8, 2009 #6


    User Avatar
    Homework Helper

    By writing [tex]\sqrt{x} = r[/tex], it means that, r is the principal square root of x.

    The principal square root of a non-negative number x is the number that satisfies the 2 following properties:
    • Firstly, it's a non-negative number, i.e r >= 0. (1)
    • And secondly, its square is x (i.e, it's one of the 2 square roots of x). (2)

    For any positive real x (i.e x > 0), there are 2 distinct square roots of x:
    • The principal one: [itex]\sqrt{x} > 0[/itex]
    • And the other one: [itex]-\sqrt{x} < 0[/itex]

    For x = 0, the 2 square roots becomes 1, which is: [tex]\sqrt{0} = 0[/tex]

    For x < 0, there's no real square root of it. There exists complex ones, but just don't worry about it for now. :)

    [tex]\sqrt{(-4) ^ 2} \neq -4[/tex] because -4 is negative, and hence cannot be the principal square root of (-4)2.


    Using the 2 properties of the principle square root above, namely (1), and (2), we derive the following formula:
    [tex]\sqrt{\alpha ^ 2} = |\alpha|[/tex].

    Which can be easily proven. You can give it a try, if you want. :)
  8. Sep 8, 2009 #7
    In that case I agree with you, but when I replied it was already [tex]\sqrt{(-4)^2}[/tex]. Perhaps I should've read your post more carefully.
    Last edited by a moderator: May 4, 2017
  9. Sep 8, 2009 #8


    Staff: Mentor

    |-4| [itex]\neq \pm 4[/itex]! The absolute value of a number has a single value that is always nonnegative!
    This doesn't make any sense. For real numbers, the square root function's domain is nonnegative real numbers. The number 4 does not have a negative value.
  10. Sep 8, 2009 #9
    Sorry about it I meant |-4|=|[itex]\pm 4[/itex]|=4. I will fix it immediately. :eek:
  11. Sep 9, 2009 #10

    [tex]\sqrt{(-4)^2} \neq \sqrt{-1(4^2)}[/tex]

    The left side equals 4 while the right equals 4i (not 4). [itex](-4)^2 = (-1 \cdot 4)^2 = (-1)^2(4)^2 = (1)(16) = 16.[/itex]

    In general

    [tex]\sqrt{x^2} = \left| x \right|[/tex] as has been mentioned.

    Last edited: Sep 9, 2009
  12. Sep 9, 2009 #11


    Staff: Mentor

    In addition to Elucidus's comment, there is a factoring property for the square root (and other roots) that is often misused.

    If a and b are nonnegative, then [itex]\sqrt{ab}~=~\sqrt{a}\sqrt{b}[/itex].
  13. Sep 9, 2009 #12
    Notice that he wrote


    instead of


    Thou shalt not misquote.
  14. Sep 9, 2009 #13
    Indeed, I mistook that quote. Apologies. The OP stated [tex]\sqrt{(-4)^2}[/tex] and I mistakingly thought jhooper3581 was citing that expression. He did not and I missed that.

  15. Sep 12, 2009 #14
    So did I at first.
  16. Sep 12, 2009 #15
    So would it be correct to say that although there is a positive and a negative square root to a number, we only consider the positive one since the square root is a function and functions can only have one y value for every x value?
  17. Sep 12, 2009 #16
    No that is simply not correct. Consider y = x2. In fact, this is your problem.

    Here is a illustration. I hope that after all those posts you will understand what these people are saying to you.

    http://img225.imageshack.us/img225/42/graphofquadratic.png [Broken]
    Last edited by a moderator: May 4, 2017
  18. Sep 12, 2009 #17

    Actually I believe I am correct, well at least to some degree. I think that you just misunderstood me. I said that since the square root is a function, you can only have one y value for each x value, thus we only consider the principal square root. What are you trying to show with your illustration?
    Last edited by a moderator: May 4, 2017
  19. Sep 12, 2009 #18
    Ok, if your function is [itex]y=\sqrt{x}[/itex], then there is unique value for y, by choosing different [itex]x \in [0, +\infty] [/itex]. And do you know what makes the function with that property?

    The function is neither odd nor even.
  20. Sep 12, 2009 #19


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    Homework Helper

    Well, what do you mean? Which function are you referring to? Of course the square root function is neither odd, nor even, because its domain is R+, which is clearly not symmetric to 0. But I'm wondering if it does have anything to do here?!?!

    As far as my knowledge goes, this is correct. :)
  21. Sep 12, 2009 #20
    Thanks VeitDao, you've been a great help.
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