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Square Root, +/- sign?

  1. Jul 13, 2009 #1
    Ok, I know this question sounds incredibly elementary, but please don't just dismiss it, try to understand what I'm REALLY asking.

    Ok, say we have [itex]x^{2} = 9[/itex]. I know the answer is [itex]x = \pm 3[/itex].
    But I was just thinking about the general rule to this. I was thinking that what really happens is this first:
    [itex]\pm x = \pm 3[/itex]
    Which can then reduce to:
    [itex]x = \pm 3[/itex]
    so the general rule is that anytime you take something out of a square root, it gets a +/- sign.
    Obviously no one would write it with a +/- in front of the x, it can be absorbed into the +/- on the other side. So please, don't say the step is "not needed" or "redundant" because I agree with you. I'm simply asking this because if it's not the case that there is a general rule like that, then we must come to the conclusion that there are certain "situations" in which you add it and sometimes you don't. I have been thinking about it, and I cannot see a situation in which writing the +/- on both sides is invalid. It always works, so why can't we say that this is what really goes on behind the scenes?

    Sorry if this seems longer than it should be but I am in an argument with someone about this at the moment so I'm making my case.
    For example:
    say we know: x = 5
    and we have this relation [itex]x^{2} = 25[/itex]
    in this situation we have two solutions 5 works, and so does -5. Therefore [itex]\pm x[/itex] is a solution.

    also keeping the +/- in front of x retains more information then just saying x alone, because if you know a solution to a problem is (+/-)x and you also know that x = 7, then you know right away that a second solution is -7.

    So back to the beginning:
    [itex]x^2 = 9[/itex] then the next step,
    [itex]\sqrt{x^{2}} = \sqrt{9}[/itex]
    now... am I to believe that there is a special rule that variables like x don't get the +/- ? Or, does it seem more general to just tack a +/- to both sides?

    I apologize for what this must sound like, but I really would just like some closure on this, and I'm actually a little embarrassed to even have to ask such a question.
    Last edited by a moderator: Jul 13, 2009
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  3. Jul 13, 2009 #2


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    I think the problem with this is that let's say you setup the plus-minus on both sides. Now, the convention is that a [tex]\pm[/tex] is consistent throughout the equation, that is if you have [tex]y = a \pm \sqrt {3x \mp 5} [/tex], you have 2 solutions: [tex]y = a + \sqrt {3x - 5} [/tex] and [tex]y = a - \sqrt {3x + 5} [/tex]. If you take the "plus" solutions with what you have, you could have something such as x=-3 be a solution to what we're looking at but an inconsistent solution to your way of doing things. You'd have a +*(-3) = +3
  4. Jul 13, 2009 #3
    I meant that you would add the [itex]\pm[/itex] sign, after you took the square root, not while it's still under the radical.
    Perhaps I just misunderstood you though.
  5. Jul 13, 2009 #4


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    I mean't [tex] y = a \pm \sqrt {3x \mp 5} [/tex] simply as an example of the convention of writing [tex]\pm[/tex]. Heck we could do [tex]y = a \pm 3x \mp 5[/tex] but that's not the point I'm making. I'm simply showing that your plus/minus has a certain convention, that is you have the "up" solution and "down" solution where you use whatevers on top or whatevers on bottom of the plus/minus. With your system it can get screwed up since if you follow the convention, you could use a valid solution such as, for your problem of the square root of 9 being -3 and 3, x=-3 and wind up with an inconsistent statement.
  6. Jul 13, 2009 #5
    The rule is [itex]\sqrt{x^{2}} = x[/itex] and it is wrong to say [itex]\sqrt{x^{2}} = \pm x [/itex]. It doesn't matter whether x is an explicit numeral like 3, or if x is a placeholder for a numeral, the rule is the same.

    Therefore when you apply the rule to [itex]\sqrt{x^{2}} = \sqrt{9}[/itex] is reduces to [itex]x = 3[/itex] according to the rule I just gave.

    Seperately we notice that [itex]x = -3[/itex] is also a solution, so the rule I gave above does not find all solutions, but it is the usual one.

    I prefer the modern computer algebra convention that [itex]\sqrt{x^{2}}[/itex] cannot be simplified until we know the domain of x, this way we do not lose solutions so carelessly.
  7. Jul 14, 2009 #6
    You problem is very early on with your statement [itex]\pmx=\pm3[/itex] that is incorrect. Simply consider the minus case, then you would have [itex]-x=-3\rightarrow-x+3=0[/itex] which has the solution x=3 not x=-3. In other words [itex](\pmx=\pm3)\neq x=\pm3[/itex].
  8. Jul 14, 2009 #7


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    If [itex]x^2= 9[/itex] then you really get four equations: x= 3, x= -3, -x= 3, and -x= -3. Of course, the first and last are equivalent as are the second and third so the reduce to x= 3 and x= -3.
  9. Jul 14, 2009 #8


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    The convention that used to be taught when solving equations like this can be summarized (approximately) as
    1) The square root of a positive number is positive
    2) the square root of a perfect square involving a variable is the absolute value of the expression

    so that

    x^2 = 9

    becomes, after "square roots",

    |x| = 3

    and the values that solve this equation are [tex] \pm 3 [/tex].
  10. Jul 14, 2009 #9


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    I get:
    [tex]x^{2}=9[/tex] which is equivalent to:
    which can be factorized as:
    yielding TWO equations,
    [tex]x+3=0[/tex] or [tex]x-3=0[/tex]
  11. Jul 14, 2009 #10


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    Yes. But I don't see how that contradicts anything I said.

    You could, as easily, have factored [itex]x^2- 9[/itex]] as (-x+ 3)(-x- 3) giving the equations -x= -3 and -x= 3 which, as I said, are equivalent to x= 3 and x= -3. This was in response to the original post in which he argued that [itex]x^2= 9[/itex] reduced to [itex]\pm x= \pm 3[/itex].

    Of all of these responses, I think statdad's that [itex]x^2= 9[/itex] reduces to |x|= 3 is best.

    Civilized, your statement that "The rule is [itex]\sqrt{x^2}= x[/itex] and it is wrong to say [itex]\sqrt{x^2}= \pm x[/itex]." is incorrect. Yes, it is wrong to say [tex]\sqrt{x^2}= \pm x[/itex] but it is also wrong to say [itex]\sqrt{x^2}= x[/itex]. What if x= -3? Then [itex]x^2= 9[/itex] and [itex]\sqrt{x^2}= \sqrt{9}= 3= -x[/itex]. What is true is, again, statdad's statement [itex]\sqrt{x}= |x|[/itex].

    But I don't interpret "If [itex]x^2= 9[/itex] then [itex]\pm x= \pm 3[/itex]" as taking the square root of both sides. Certainly [itex](x)^2= x^2[/itex], [itex](-x)^2= x^2[/itex], [itex](3)^2= 9[/itex], and [itex](-3)^2= 9[/itex] so, as I said before, all of x= 3, x= -3, -x= -3, and -x= 3 will give, by squaring both sides, [itex]x^2= 9[/itex].
    Last edited by a moderator: Jul 14, 2009
  12. Jul 14, 2009 #11
    If you put it like this: [tex]\pm x[/tex] = [tex]\pm3[/tex] doesn't that say that x = [tex]\pm3[/tex] and -x = [tex]\pm3[/tex]? That really doesn't make alot of sense. If x = 3, then -x = -3, but it can't be both simultaneously.
  13. Jul 14, 2009 #12
    It is elementary, but it's an important question that too many students dismiss, because they "already know the answer." However, as a student going from algebra or route calculus to a more flexible understanding of mathematics, this is a critical question to ask.

    Let's just restrict ourselves to integers to make our problem as simple as possible.

    Now ask the question, if [itex]x^{2} = 9[/itex], what is x?

    Your teacher claims its this thing called [itex]\pm 3[/itex]. But what is "plus or minus two"? Is it a number? I must be two numbers.... +3 and -3. But it's incorrect to say that x = +3 and x = -3, because equal things are equal, and that would imply +3 = -3.

    Well, it's plus OR minus, not plus AND minus, so it is either true that x = +3 or x = -3. That way, we don't run into that silly +3 = -3 problem. But suppose we have two integers x and y where [itex]x = \pm 3[/itex] and [itex]y = \pm 3[/itex]. Is it true that x = y? Well, it might be the case that x = 3 and y = -3, so no, not necessarily. Again, we have issues with our rules for equality.

    What you come to find out is that [itex]\pm 3[/itex] doesn't work like a number at all. It doesn't work like one, because it ISN'T one. It doesn't obey algebra, and you can easily end up with contradictions even when you follow the rules.

    In reality, [itex]\pm 3[/itex] is something we call an "abuse of notation." It's a helpful device that gets you to the right answer without being strictly valid.

    The heart of the matter is that the solution to this problem is not unique. When you say [itex]x = ...[/itex], you give x a definite, unique value. However, as we know logically, the number whose square is 9 is not a unique object.

    This is a type of problem that comes up a lot in mathematics. You start with some property. Then, you try to figure out what objects have that property. There may be a unique solution. There may be two objects that both have the property (as in this problem). It may be that this property is held by an entire class of objects. There might even be none at all.

    You can never make an assumption about the number of objects which satisfy a property. It has been an embarrassment for a few math PhD's who've spent years studying the objects with a particular property.... only to find out a year later that there are no such objects.

    So the bottom line is always be alert when you see and use [itex]\pm[/itex] in your class. Keep in mind that it's just a shorthand for a more complicated truth. When you finish with a problem, go back and reason your argument from start to finish. If you can, plug in both values and see if they both make sense. Be especially mindful of squares and division by zero.
  14. Jul 14, 2009 #13
    I'm with you man, in fact that is exactly one of the arguments I've been using. The four equations simplify to the regular two +3 and -3.

    Ok.. for those of you who think it's not the correct convention, then how would you modify this scenario?

    x > 0 (so x is some positive value, can't be negative, but we don't know what it is).
    we know that a solution is x. But there is a second solution equivalent to the negative of x.
    So in this situation you would lose information if you just said the answer is x. But then later say we define x as 5, if you wrote the simplification as (+/-)x, you will know that the solution is 5 AND -5.
    [itex]\sqrt{x^{2}} = x[/itex]
    [itex]\sqrt{x^{2}} = -x[/itex]

    Also, I'm not so sure that you have a single set of "up" and "down" solutions as Pengwuino says, I always thought it was like HallsofIvy said, you would have 4 solutions. The only time the "up" and "down" solutions apply is if you use the same variable like [itex]\pm x * \frac{1}{2}(\mp x)[/itex] or something like that. Then you only have two solutions, where the second x is the opposite sign of the first, but if they are separate variables then you have four solutions.
    I'm not saying I'm right, but that is just my guess, please let me know if that is wrong.

    Thanks for that post Tac-Tics, I think that really cleared some things up, in fact I think that's why it went from a discussion with this person to an argument, because "they already know the answer" as you said. I'm sorry if I missed it, but what is your suggestion then about the notation? Or is there just no simple answer to this question?
  15. Jul 14, 2009 #14
    You're welcome =-D

    The notation is basically insoluble. The equals sign just doesn't play nicely with [itex]\pm[/itex]. It is best to break it up into sentences. Alternatively, you could use logical notation.

    [tex]x^2 = 9[/tex] implies [tex]x = 3 \wedge x = -3[/tex]

    When you do this, you have some bookkeeping to do. From years of math classes, you become used to the idea that "anything written on the paper is true." That is, if you have "x = 20" at the top of the page and you forget about x for a while, you know that when you get to the bottom of the page, x is still equal to 20.

    However, when you have a logical or connecting two sentences, in a sense, you create two independent chains of reasoning. You might want to divide the paper in half with a line. The steps you take on the left hand side is the universe where you suppose x = -3. The steps you take on the right side of the paper is the universe where you suppose x = +3. Then, if you can manage to draw particular conclusion on BOTH sides of the page (for example, |x| = 3), you can say that that conclusion is a logical deduction based on the stuff at the very top of the page ([tex]x^2 = 9[/itex])
  16. Jul 14, 2009 #15


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    No, it means [tex]x= \pm 3[/tex] or [tex]-x= \pm 3[/tex] not "and"!
  17. Jul 14, 2009 #16
    Ok, well then revising the question a little in light of what Tac-Tics is saying, wouldn't you still split into two alternate lines of reasoning when you see this:
    where one path assumes the +x solution, and another path assumes the -x solution. Right?
    So that would still be similar to what I was asking at the beginning with x^2 = 9. You split up into two paths with +x and -x, as well as two paths for 3 and -3, giving (x=3, x=-3, -x=3, -x=-3) with (as HallsofIvy stated) two of the paths being equivalent so we end up with simply two alternate lines of reasoning in which (x=3, x=-3)

    Does that sound good to everyone?
  18. Jul 14, 2009 #17
    This discussion is nothing but technical overkill... Plus/Minus sign is JUST a shorthand notation.

    In case of doubt, use the explicit values. +3,-3 and you are done.

    I don't agree about having 4 equations reduced to 2. Because one can go for infinite solutions just as innocent as above, by using dummy relations to both sides of the equations.
  19. Jul 15, 2009 #18
    It is just a shorthand. The problem is that that point is rarely made explicit. Lots of algebra students wind up pretending "plus or minus three" is an actual number, trying to manipulate it in symbols, and examining its meaning only at the end of the problem.

    The thing is this approach usually works fine. It isn't until later that it blows up. The teacher gets to problems where you divide by the value x - 3 early on, then later in the same problem, you determine that [tex]x = \pm 3[/tex]. The students are then angry to get marked down because [tex]x = +3[/tex] leads to a hidden division by 0.
  20. Jul 15, 2009 #19
    Yes trambolin. But what I am arguing for is that taking the square root of anything leads to a positive and negative solution. So what I am trying to say with the example x^2 = 9 is that what is really happening is that you take the square root on each side individually, you don't look at the whole thing, and if this is true then you should have [itex]\pm x = \pm 3[/itex]
    which is exactly the same thing as the common form:
    [itex]x = \pm 3[/itex] (please understand that I am NOT saying people should write (+/-)x = (+/-)3, I agree it is completely redundant, I am simply showing that that is the first step which people usually skip).
    or however you want to represent them. and I do mean + OR -, I'm pretty sure that I am not confused about this.

    So because you can always take the + and - solution of both sides and it reduces to the same thing that everyone is used to seeing.. Then why is it so hard to believe that what I am describing is simply the general rule.
    I find it hard to believe that the geniuses coming up with this stuff said "ok so when you take the square root, you just have to think about it and realize that 'sometimes' you have two solutions". To me it seems a lot more satisfying that you just always have two solutions, but further reasoning can be used to reduce this, it seems much more complete, and fundamental like a mathematical operator should be.

    This is what I was trying to say with my first post, I'm not very good at explaining things sometimes.
    Last edited by a moderator: Jul 15, 2009
  21. Jul 15, 2009 #20
    The universal convention for square roots is that [tex]\sqrt{x} \geq 0[/tex] for all x. That is, the square root always provides you with the positive result, never the negative one. If you want the negative one, you simply ask for [tex]-\sqrt{x}[/tex] instead.
  22. Jul 15, 2009 #21
    OK, maybe I should write slower...

    Here is the point where I don't agree. When you take the square root of x^2 = 9, you can not write [itex]\pm x = \pm 3[/itex] Because one side of the equation is the variable that you want to assign values to. Hence there is no ambiguity about the variable's sign. What we are trying to express mathematically is that there are two values that satisfy this equation. Taking a square root is just a multivalued mapping. But putting (+/-) is not an mathematical operator. It is a conventional expression. And this is the most compact form that people came up with, so far. But if math rigor is the religion, we mean [itex]x\in\{-3,3\}[/itex]

    Maybe this example shows better what I am trying to emphasize. For example if I write [itex]25 = 25 \Longrightarrow\pm 5 = \pm 5[/itex] everyone including you will dismiss that immediately. Because there is no such thing. Your extra freedom comes from the fact that one side is a variable and can not dictate sign information.
  23. Jul 15, 2009 #22
    I'm not saying that (+/-) x dictates sign information. All it means is that x (whether it be positive or negative) is a solution, but so too is the opposite sign of x (therefore we write -x), I'm not saying x is negative, only that the negative of x also satisfies the equation sqrt(x^2). Also, what if x does have a definite value, we just don't know it, but we know x>0. Then purely mathematically we have two solutions to [itex]\sqrt{x^{2}}[/itex] which is x, and the opposite sign of whatever x is.

    Also, as I was saying earlier that if we know the variable is the same then it doesn't treat them separately, meaning [itex]\pm 5 = \pm 5[/itex] would be the same as two solutions in this case, not four, we would have 5 = 5, and -5 = -5 because it's the same value or variable (if we used one). So this simply reduces to one solution 5 = 5.

    So for instance if we had [itex]\sqrt{x^{2}} = \sqrt{y^{2}}[/itex] then we would have [itex]\pm x = \pm y[/itex] which is exactly the same as [itex]x = \pm y[/itex] (or vice versa).
    Then later if we know that x = y. Then the negative line of reasoning x = -y becomes invalid because we now treat them as the same variable, which is the reason we don't end up with something like 5 = -5.

    Again, I am not writing this to "prove" that I am right, I'm just saying my thoughts and hoping you all can help me to know if they are right or wrong, so even though I say something like "then the negative line of reasoning x = -y becomes invalid" I am just guessing and would completely understand if something I say is just flat out wrong.
  24. Jul 15, 2009 #23


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    "Here is the point where I don't agree. When you take the square root of x^2 = 9, you can not write Because one side of the equation is the "

    As noted earlier, when you take the square root you should write

    |x| = 3


    1) since [tex] x [/tex] is a variable, without prior information you don't know its sign, but the basic [tex] \sqrt{} [/tex] expression always returns a non-negative result
    2) by definition of the square root function, [tex] \sqrt 9 = 3 [/tex]

    Solving [tex] | x | = 3 [/tex] gives the two solutions.

    Lest tempers begin to flare over this, a (humorous) little story should be considered (look at the final line: let's not get to that point):

    "Standard mathematics has recently been rendered obsolete by the discovery that for years we have been writing the number five backwards. This has led to a re-evaluation of counting as a method of getting from one to ten. Students are taught advanced concepts of Boolean Algebra, and formerly unsolvable equations are dealt with by threats of reprisals." (Woody Allen, in the book Getting Even )
  25. Jul 16, 2009 #24
    I am not getting sensititive or trying to prove you wrong. I am just saying the discussion on this issue is a re-wrapping of the original concept, hence can be deduced from the very same amount of information. That is why I started by saying (Maybe a little bit, in a rude fashion) this is a technical overkill.

    Also considering [itex]\sqrt{x^{2}} = \sqrt{y^{2}}[/itex], concluding that [itex]\pm x = \pm y[/itex] mathematically enforces that [itex]x=y= 0[/itex]. Hence it is not the same thing by writing [itex]x=\pm y[/itex]
  26. Jul 16, 2009 #25


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    "That is why I started by saying (Maybe a little bit, in a rude fashion) this is a technical overkill."

    I was agreeing.

    "Also considering [tex] \sqrt{x^2} = \sqrt{y^2}[/tex] , concluding that mathematically enforces that x = y = 0."

    Here i do not agree: from [tex] \sqrt{x^2} = \sqrt{y^2} [/tex] you can conclude that this equation: [tex] |x| = |y| [/tex] is satisfied, but assigning specific values past that requires more information.
    Here I disagree:
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