anonymous
  • anonymous
Can someone explain why |x| = sqrtx^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\sqrt{9} = \pm3\]"sqrt of positive is also always positive" ???
experimentX
  • experimentX
sorry about that
anonymous
  • anonymous
so, anyone???

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experimentX
  • experimentX
then the above relation is not true
experimentX
  • experimentX
i mean not absolutely
anonymous
  • anonymous
thats what i think, but someone just said it is, turinTest, i think. And he was preaty sure about it....
experimentX
  • experimentX
can you show me where??
experimentX
  • experimentX
√9 !=-3 but √9=+3
anonymous
  • anonymous
ya, i just so his explanation, but loks so forced to me. Maybe just my problem
experimentX
  • experimentX
yeah that's right, because he is taking the positive root of x, which is equal to the positive value of x
experimentX
  • experimentX
but x can one value between -ve and +ve, but absolute value of x will always be equal to the positive root of x^2
anonymous
  • anonymous
\[|x|=-x (x<0)\]\[|x| = x (x>0)\] so, to puting them bouth together he using |x| for root values
anonymous
  • anonymous
root is multivalued function...
experimentX
  • experimentX
to be honest I made silly statement with out a thought. 'root of positive is always positive' ... but i should rather have been 'positive root of ..'
anonymous
  • anonymous
right
anonymous
  • anonymous
When you square a number, you lose the sign. Once it is gone, it doesn't come back. So taking the root of the square is just like taking the absolute value.
experimentX
  • experimentX
he is taking THE POSITIVE ROOT
experimentX
  • experimentX
heheh ... positive root will always be positive and negative root will always be negative. in above case you are also taking the positive root
anonymous
  • anonymous
hmm.. I agree with sign los, but when you take the root you will get +/- anyway
experimentX
  • experimentX
and also -|x| = - sqrtx^2
anonymous
  • anonymous
right
anonymous
  • anonymous
so i guess:\[|x| = \sqrt{x ^{2}}\] is just the way to write: \[\pm x = \sqrt{x ^{2}}\]
experimentX
  • experimentX
no ... the second is not right, because on the right hand side ... you have already chosen the positive root
anonymous
  • anonymous
i don't think i chouse anything. x can be negative, can be positive. Same for the sqrt
experimentX
  • experimentX
the EXPRESSION on right hand side must be positive
anonymous
  • anonymous
x2 is >0 no mater what x is
experimentX
  • experimentX
yup, that is if x has real value
anonymous
  • anonymous
sorry, my browser won't let me use the equation applet, but: (the square root of x) squared is always going to be positive
anonymous
  • anonymous
and why? plz explain
anonymous
  • anonymous
any number squared will be positive, no?
experimentX
  • experimentX
because if x has a real value, then the value of x is either going to be positive or negative .. and the positive multiplication with positive is always positive and -x- is also always postive
anonymous
  • anonymous
if real yes. But continue
experimentX
  • experimentX
But??
anonymous
  • anonymous
I have no imagination. Nor do I understand sarcasm. Sorry.
anonymous
  • anonymous
my point is: you "aplye" srt function to a square of a nummber. Root function gives you two possible values +/-. You don't know what x (+,-) was used to get the square.
experimentX
  • experimentX
the root function doesn't give you two value ... the root function gives you one value ... it's only becuse you have two roots, one negative the other positve
experimentX
  • experimentX
and also be definition of function, function gives you one value ..
anonymous
  • anonymous
there are multivalued functions, and square root is one of them...
anonymous
  • anonymous
arsen another
anonymous
  • anonymous
arcsen*
experimentX
  • experimentX
let me repeat this, a function CANNOT give you two values,
anonymous
  • anonymous
it can and it gives
experimentX
  • experimentX
and arcsin(x) is not a function, if it's range is not defined with in [-pi/2, pi/2]
experimentX
  • experimentX
In mathematics, a function[1] is a correspondence from a set of inputs to a set of outputs that associates each input with exactly one output --- from wikipedia
anonymous
  • anonymous
a function is a set of ordered pairs of one, or two sets. No more restrictions are aplyed
AccessDenied
  • AccessDenied
a function maps each value in its domain to one not necessarily unique value of the range you will not have a function by definition with multiple y-values for one x-value
experimentX
  • experimentX
no .. not exactly, a function must give exactly one output for one or many input // as accessdenied said, unique is not necessary condition ... but it must give one output ... not many outputs
anonymous
  • anonymous
agree about that each value of the domain is mapped. But can be mapped to more than one value in codomain
anonymous
  • anonymous
sqrt of 1 has 2 values. Third root: 3, and so on
anonymous
  • anonymous
nth root has n values
experimentX
  • experimentX
no .. sqrt has one and only one value X^2 - 1 = 0, x=+-1 are roots of equation on/// not the value of srt 1
anonymous
  • anonymous
sqrt has two values. The positive one is cold principal value, but there is also a negative one.
experimentX
  • experimentX
can you show me in graph???
experimentX
  • experimentX
f(x) = sqrt(x)
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Multivalued_function look there in examples for multivalued functions. For graphing you need to deside which value will you use, normaly principal is taken
anonymous
  • anonymous
sqrt graph maps positive R to it self, but that principal value of the root
experimentX
  • experimentX
multivalued function and function are different things ... they have different criteria and different things ,, the graph of sqrt x is given here http://www.wolframalpha.com/input/?i=%28x%29%5E1%2F2
anonymous
  • anonymous
i know the graph, thats not the point of my question
experimentX
  • experimentX
if sqrt x is a multivariable function then it must be something like this |dw:1332703248590:dw|
anonymous
  • anonymous
a function like you say is so called "well defind function". It can be graphed. Multivalued function not always can be graphed entirly, its used to study relations between sets. My question came from the use of this |x| in a study of limits, that not necesarly have to be related to graph. Its more about sequences, so thats my point.
anonymous
  • anonymous
by the way, your graph is wrong
anonymous
  • anonymous
|dw:1332703451869:dw|
experimentX
  • experimentX
what ever it is ... my point is at any particular point x, if the relation is a function then, it must have only have one y.
AccessDenied
  • AccessDenied
From what I had learned, the symbol itself for square root only indicates the principal root... if you're only referring to the principal root, then the x^2 would still become |x| because we have to account for the fact that x can be positive or negative to get the same result of x^2... at least, that's how I am looking at it.
anonymous
  • anonymous
agree with you Access, thanks for helping
anonymous
  • anonymous
access is right, i think. The symbol meaning is the point here.
experimentX
  • experimentX
i quite disagree ... square root and roots of quadratic eqn are two different things, x^2 - 4 =0 is a quadratic eqn with roots +2 and -2 while, sqrt(4) is a complete different thing
experimentX
  • experimentX
sqrt(4) = 2 and only 2 NOT -2 while, x = both, and |x| = 2
anonymous
  • anonymous
de definition of sqrta is: b2 =a, and also u see sometimes: b = +-sqrta
anonymous
  • anonymous
so looks like the symbol it self is considered positive
anonymous
  • anonymous
ok, thx for help
experimentX
  • experimentX
you are welcome
AccessDenied
  • AccessDenied
The idea of the square root is answering the question: "y^2 = a, what is y?" There would be two answers to this question, a positive and a negative valuie. If we use the mathematical representation of square root, "sqrt" (this is principal root), we would say y=sqrt(a) and y=-sqrt(a). You can see this is true by substituting: (sqrt(a))^2 = a, and (-sqrt(a))^2 = (-1)^2(sqrt(a))^2 = a. Both will work. Glad to help. :)

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