Can someone explain why |x| = sqrtx^2

- anonymous

Can someone explain why |x| = sqrtx^2

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- anonymous

\[\sqrt{9} = \pm3\]"sqrt of positive is also always positive" ???

- experimentX

sorry about that

- anonymous

so, anyone???

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- experimentX

then the above relation is not true

- experimentX

i mean not absolutely

- anonymous

thats what i think, but someone just said it is, turinTest, i think. And he was preaty sure about it....

- experimentX

can you show me where??

- experimentX

√9 !=-3 but √9=+3

- anonymous

ya, i just so his explanation, but loks so forced to me. Maybe just my problem

- experimentX

yeah that's right, because he is taking the positive root of x, which is equal to the positive value of x

- 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

\[|x|=-x (x<0)\]\[|x| = x (x>0)\]
so, to puting them bouth together he using |x| for root values

- anonymous

root is multivalued function...

- 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

right

- 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

he is taking THE POSITIVE ROOT

- 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

hmm.. I agree with sign los, but when you take the root you will get +/- anyway

- experimentX

and also -|x| = - sqrtx^2

- anonymous

right

- anonymous

so i guess:\[|x| = \sqrt{x ^{2}}\]
is just the way to write:
\[\pm x = \sqrt{x ^{2}}\]

- experimentX

no ... the second is not right, because on the right hand side ... you have already chosen the positive root

- anonymous

i don't think i chouse anything. x can be negative, can be positive. Same for the sqrt

- experimentX

the EXPRESSION on right hand side must be positive

- anonymous

x2 is >0 no mater what x is

- experimentX

yup, that is if x has real value

- 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

and why? plz explain

- anonymous

any number squared will be positive, no?

- 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

if real yes. But continue

- experimentX

But??

- anonymous

I have no imagination. Nor do I understand sarcasm. Sorry.

- 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

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

and also be definition of function, function gives you one value ..

- anonymous

there are multivalued functions, and square root is one of them...

- anonymous

arsen another

- anonymous

arcsen*

- experimentX

let me repeat this, a function CANNOT give you two values,

- anonymous

it can and it gives

- experimentX

and arcsin(x) is not a function, if it's range is not defined with in [-pi/2, pi/2]

- 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

a function is a set of ordered pairs of one, or two sets. No more restrictions are aplyed

- 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

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

agree about that each value of the domain is mapped. But can be mapped to more than one value in codomain

- anonymous

sqrt of 1 has 2 values. Third root: 3, and so on

- anonymous

nth root has n values

- 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

sqrt has two values. The positive one is cold principal value, but there is also a negative one.

- experimentX

can you show me in graph???

- experimentX

f(x) = sqrt(x)

- 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

sqrt graph maps positive R to it self, but that principal value of the root

- 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

i know the graph, thats not the point of my question

- experimentX

if sqrt x is a multivariable function then it must be something like this |dw:1332703248590:dw|

- 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

by the way, your graph is wrong

- anonymous

|dw:1332703451869:dw|

- 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

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

agree with you Access, thanks for helping

- anonymous

access is right, i think. The symbol meaning is the point here.

- 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

sqrt(4) = 2 and only 2 NOT -2
while, x = both, and |x| = 2

- anonymous

de definition of sqrta is: b2 =a, and also u see sometimes: b = +-sqrta

- anonymous

so looks like the symbol it self is considered positive

- anonymous

ok, thx for help

- experimentX

you are welcome

- 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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