Square Root, +/- sign?

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

 

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.

 

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.

 

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].

 

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.

 

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

[tex]
x^2 = 9
[/tex]

becomes, after "square roots",

[tex]
|x| = 3
[/tex]

and the values that solve this equation are [tex] \pm 3 [/tex].

 

???????

I get:
[tex]x^{2}=9[/tex] which is equivalent to:
[tex]x^{2}-9=0[/tex]
which can be factorized as:
[tex](x+3)(x-3)=0[/tex]
yielding TWO equations,
[tex]x+3=0[/tex] or [tex]x-3=0[/tex]

 

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].

 

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.

 

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.

 

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])

 

No, it means [tex]x= \pm 3[/tex] or [tex]-x= \pm 3[/tex] not "and"!

 

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.

 

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.

 

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.