20190927, 21:20  #1 
Jul 2014
2×3^{2}×5^{2} Posts 
Can some explain this?
I don't understand why \( \sqrt{ab}=\sqrt{a}\sqrt{b} \) fails for negative a and b.
e.g \( \sqrt{(4)(9)}=\sqrt{36}=6\neq\sqrt{(4)(9)}=\sqrt{(1)(4)(1)(9)}=\sqrt{1}\sqrt{1}\sqrt{4*9}=1\times6 \) Any replies helping me understand this? Last fiddled with by wildrabbitt on 20190927 at 21:21 
20190927, 21:32  #2 
"Curtis"
Feb 2005
Riverside, CA
2^{5}×3^{2}×19 Posts 
The square root of x^2 is x, not x.
So the square root of (1)^2 is 1, not 1. It's not OK to write sqrt(x^2) = sqrt(x) * sqrt (x) = x, but that's what you did for 1. 
20190927, 21:48  #3 
Jul 2014
2·3^{2}·5^{2} Posts 
Thanks very much. I feel relieved.

20190928, 01:50  #4  
Romulan Interpreter
"name field"
Jun 2011
Thailand
2×7^{2}×103 Posts 
Quote:
\( \sqrt{(4)\times(9)}=\sqrt{36}=\pm 6=\sqrt{(4)\times(9)}=\sqrt{(1)\times(4)\times(1)\times(9)}=\sqrt{1}\times\sqrt{1}\times\sqrt{4\times 9}=\pm i\times\pm i\times\pm 6 =\pm 6 \) What you did is like saying that sqr fails (in general) because \(\sqrt 4=2\ne\sqrt 4=2\) Last fiddled with by LaurV on 20190928 at 01:52 

20190928, 02:14  #5 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}·17·97 Posts 

20190928, 02:30  #6 
"Curtis"
Feb 2005
Riverside, CA
2^{5}·3^{2}·19 Posts 
In every precalc & calculus curriculum I've seen, square root is a function. That means it has one solution, the positive one. Consider a graph of y = sqrt(x); it's the top arm of a sideways parabola, not both arms.
Contrast that with the action of taking the square root of both sides of an equation (say, to remove a power of 2 from a variable). That action leads to two solutions, a positive sqrt and a negative sqrt. x^2 = 9 has two solutions, but sqrt(9) = 3. Last fiddled with by VBCurtis on 20190928 at 02:31 
20190928, 13:42  #7 
Jul 2014
2·3^{2}·5^{2} Posts 
I knew the definition of root x as a function is the positive root but still I was bamboozled by the statement I wrote (I read it somewhere).
Is the same true for \(x^4\)? 1^4 = 1 = 1 \(\sqrt[4]{((\sqrt{2})(\sqrt{2})(\sqrt{2})(\sqrt{2}))}=\sqrt{2} = \sqrt{2}\) ? 
20190928, 15:03  #8 
Dec 2012
The Netherlands
5·353 Posts 
If x is a real number, then \(x^4\) is easier because \(x^2\geq 0\) so \(\sqrt{x^4}=\sqrt{(x^2)^2}=x^2=x^2\)
and hence \(\sqrt[4]{x^4}=\sqrt{x^2}=x\). 
20190928, 15:11  #9 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}·17·97 Posts 

20190928, 17:40  #10 
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
11,491 Posts 
Hmm?
(2i)^4 = ((2i)^2)^2 = (4)^2 = 16. (2i)^4 = ((2i)^2)^2 = (4)^2 = 16. ITYM, 16^(1/4) = ยฑ2, ยฑ2i. More generally, there are n distinct nth roots of unity. Ah, the OP is interested only in real roots. Last fiddled with by xilman on 20190928 at 17:43 Reason: Added final sentence. 
20190929, 02:00  #11 
Feb 2017
Nowhere
1011101011010_{2} Posts 
The fundamental issue with "wrong square root" difficulties is, in order to define noninteger powers, you need a logarithm. With a logarithm in hand, you can say
for any complex number k. For roots of positive real numbers a, and real values of k, everything is fine  you've got a realvalued logarithm, defined by an integral. You can not, of course, define a logarithm of 0. For nonzero complex numbers other than positive real numbers, though (including negative real numbers), defining a logarithm is a real problem. For one thing, the logarithm of any such number will have a nonzero imaginary part. But the real problem is, once you allow complex exponents, you are stuck with the fact that the exponential function is periodic, because so "the" logarithm is only determined up to an integer multiple of 2*pi*i. This difficulty is usually dealt with by making a "branch cut" emanating from 0, which excludes closed curves with 0 (the origin) inside them. Different integer multiples of 2*pi*i give different "branches" of the logarithm. And unless the imaginary part of every logarithm in sight is exactly zero, there is every chance that applying the usual "laws of logarithms/exponents/roots" will put you on a different "branch" of the logarithm than the one you started with. And, if you get on the wrong branch of the logarithm, you make a monkey of yourself! 
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