square root equal to a negative

LaurV · 2013-02-12, 15:46

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?

akruppa · 2013-02-12, 16:00

It does not. The two (Edit: complex) solutions are ±2i.

Dubslow · 2013-02-12, 16:01

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.

LaurV · 2013-02-12, 16:04

Quote:

Originally Posted by akruppa

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!)

LaurV · 2013-02-12, 16:06

Quote:

Originally Posted by Dubslow

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.

Dubslow · 2013-02-12, 16:11

Quote:

Originally Posted by LaurV

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

you too.

?

You pose an interesting problem.

LaurV · 2013-02-12, 16:15

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?

Dubslow · 2013-02-12, 16:31

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?

bsquared · 2013-02-12, 16:33

x = 16*i^4

NBtarheel_33 · 2013-02-12, 16:41

Quote:

Originally Posted by LaurV

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.

akruppa · 2013-02-12, 17:14

Quote:

Originally Posted by LaurV

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.

2013-02-12, 15:46	#1
LaurV Romulan Interpreter "name field" Jun 2011 Thailand 7·1,423 Posts	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?

2013-02-12, 16:00	#2
akruppa "Nancy" Aug 2002 Alexandria 2467₁₀ Posts	It does not. The two (Edit: complex) solutions are ±2i. Last fiddled with by akruppa on 2013-02-12 at 16:00

2013-02-12, 16:01	#3
Dubslow Basketry That Evening! "Bunslow the Bold" Jun 2011 40<A<43 -89<O<-88 3×29×83 Posts	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

2013-02-12, 16:31	#8
Dubslow Basketry That Evening! "Bunslow the Bold" Jun 2011 40<A<43 -89<O<-88 3×29×83 Posts	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

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Square Root Days	davar55	Lounge	0	2016-03-16 20:19
NFS Square root problems	paul0	Factoring	10	2015-01-19 12:25
square root crash	bsquared	Msieve	17	2010-04-09 14:11
Square root of 3	Damian	Math	3	2010-01-01 01:56
Divisible up to Square Root	davar55	Puzzles	3	2007-09-05 15:59

2013-02-12, 16:15	#7
LaurV Romulan Interpreter "name field" Jun 2011 Thailand 7×1,423 Posts	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?

2013-02-12, 16:33	#9
bsquared "Ben" Feb 2007 3617₁₀ Posts	x = 16*i^4