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Old 2013-02-12, 15:46   #1
LaurV
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Default square root equal to a negative

Now I have a big problem...

I have the next equation: sqrt x =-4

Does it, or does it not have a (real/integer) solution?
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Old 2013-02-12, 16:00   #2
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It does not. The two (Edit: complex) solutions are ±2i.

Last fiddled with by akruppa on 2013-02-12 at 16:00
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Old 2013-02-12, 16:01   #3
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No, its only solutions are complex (with non zero imaginary part). Anyone who claims a real solution forgets that squaring both sides introduces extra solutions which do not satisfy the original equation.

Edit: akruppa's solutions are not, I think he read the question too quickly.

Last fiddled with by Dubslow on 2013-02-12 at 16:05
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Old 2013-02-12, 16:04   #4
LaurV
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Quote:
Originally Posted by akruppa View Post
It does not. The two (Edit: complex) solutions are ±2i.
I think you confuse with a radical of a negative number, which I did not ask for. sqrt(+/-2i) in complex, is 1+/-i, and not -4.


(Gotcha!)
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Old 2013-02-12, 16:06   #5
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Quote:
Originally Posted by Dubslow View Post
No, its only solutions are complex (with non zero imaginary part). Anyone who claims a real solution forgets that squaring both sides introduces extra solutions which do not satisfy the original equation.

Edit: akruppa's solutions are not, I think he read the question too quickly.
you too.
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Old 2013-02-12, 16:11   #6
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Quote:
Originally Posted by LaurV View Post
I think you confuse with a radical of a negative number, which I did not ask for. sqrt(+/-2i) in complex, is 1+/-i, and not -4.


(Gotcha!)
Not quite. sqrt(2i) = -/+1 -/+ i , and sqrt(-2i) = +/-1 -/+ i, i.e. you forgot half the solutions.
Quote:
Originally Posted by LaurV View Post
you too.
?

You pose an interesting problem.

Last fiddled with by Dubslow on 2013-02-12 at 16:12 Reason: Clarification
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Old 2013-02-12, 16:15   #7
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Story is like that, this IS homework. My daughter's. She had to solve a system which came to this, after a lot of calculus. She said is no solution. I said the solution is 16. Indeed, sqrt(16), according with the math I know 40 years ago, is +/-4. We argued, we looked in the book (answers section). They said the solution is the empty set. I got angry and looked into wiki, and wolfram alfa, they all agree with my daughter , and I am getting crazy... We live the days when the egg teaches the hen, you know?
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Old 2013-02-12, 16:31   #8
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Hmmm... thinking of \sqrt x as a function, then it has no solutions (sqrt(16)=4). If you prefer to think of sqrt(x) as the set of all (complex) numbers whose square is x, then *maybe* you might say sqrt(16)={4,-4}.

Edit: What was the original context of this equation? As in, what was the homework question?

Last fiddled with by Dubslow on 2013-02-12 at 16:35
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Old 2013-02-12, 16:33   #9
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x = 16*i^4
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Old 2013-02-12, 16:41   #10
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Quote:
Originally Posted by LaurV View Post
Now I have a big problem...

I have the next equation: sqrt x =-4

Does it, or does it not have a (real/integer) solution?
I would say that it probably depends on how the actual symbol sqrt x is defined *within your daughter's textbook*. (Please note that this might not be consistent with what you, or I, or other advanced mathematicians know about the definition of square root, or the radical sign, but there are often (usually not-so-great) simplifications made at the elementary levels).

For instance, in my first algebra textbook, I was taught that any positive real number x had two square roots, and to answer a question like "what is the square root of 36?" I should write \pm 6. If I were solving an equation such as x^2 + 6x + 9 = 16, I would perform the steps (x + 3)^2 = 16, hence x + 3 = \pm 4, hence  x \in \{-7,  1\}.

Needless to say, it was a little embarrassing and cost me a few points on some homework assignments when I got into college, where the definition of the radical sign was taken to mean "the positive square root".

But you can see now where your daughter's question might be entirely valid and have a solution in one setting, but not in another. If, as I suspect, her present textbook is making use of the "plus-minus" definition of square roots, then one could say that, yes, if sqrt x = -4, then x = 16. On the other hand, if her textbook has taken on the more advanced, "principal"/positive-square-root-only definition, then there would be an argument that there is no solution, as there is no possible way *under the understood definition of square root *as it is in this textbook** for the square root of a number to be negative. Hence the solution would be undefined.

This, of course, is an excellent time to have a chat with your daughter (especially if she plans on going higher into math) about why this matter is an issue in the first place, and why we might like to fix "square root" to mean "only the positive square root" (e.g. to avoid a multi-valued function, to be able to take limits in calculus, to not have a weird function that raises a positive number to a positive power and gives a negative answer...) Basically, this is one of her first lessons that what works well in algebra does not necessarily work well in calculus, and why certain changes to definitions and constructs have to be made in order to fix these issues.

Last fiddled with by NBtarheel_33 on 2013-02-12 at 16:44
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Old 2013-02-12, 17:14   #11
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Quote:
Originally Posted by LaurV View Post
I think you confuse with a radical of a negative number, which I did not ask for. sqrt(+/-2i) in complex, is 1+/-i, and not -4.

(Gotcha!)
*facepalm* Indeed you did.
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