mersenneforum.org square root equal to a negative
 Register FAQ Search Today's Posts Mark Forums Read

 2013-02-12, 15:46 #1 LaurV Romulan Interpreter     Jun 2011 Thailand 3×13×229 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 246410 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
2013-02-12, 16:04   #4
LaurV
Romulan Interpreter

Jun 2011
Thailand

3×13×229 Posts

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!)

2013-02-12, 16:06   #5
LaurV
Romulan Interpreter

Jun 2011
Thailand

893110 Posts

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.

2013-02-12, 16:11   #6
Dubslow

"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

3×2,399 Posts

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.

Last fiddled with by Dubslow on 2013-02-12 at 16:12 Reason: Clarification

 2013-02-12, 16:15 #7 LaurV Romulan Interpreter     Jun 2011 Thailand 213438 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, 16:31 #8 Dubslow Basketry That Evening!     "Bunslow the Bold" Jun 2011 40
 2013-02-12, 16:33 #9 bsquared     "Ben" Feb 2007 D0D16 Posts x = 16*i^4
2013-02-12, 16:41   #10
NBtarheel_33

"Nathan"
Jul 2008
Maryland, USA

100010110012 Posts

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.

Last fiddled with by NBtarheel_33 on 2013-02-12 at 16:44

2013-02-12, 17:14   #11
akruppa

"Nancy"
Aug 2002
Alexandria

246410 Posts

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.

 Similar Threads Thread Thread Starter Forum Replies Last Post davar55 Lounge 0 2016-03-16 20:19 paul0 Factoring 10 2015-01-19 12:25 bsquared Msieve 17 2010-04-09 14:11 Damian Math 3 2010-01-01 01:56 davar55 Puzzles 3 2007-09-05 15:59

All times are UTC. The time now is 16:08.

Tue Nov 24 16:08:12 UTC 2020 up 75 days, 13:19, 4 users, load averages: 1.68, 1.75, 1.72