Square root x^2 = Absolute value x. Explain please (1 Viewer)

[Alext_]

I don't really understand this identity...can someone explain thoroughly with examples?
Cheers

 

[Kurosaki]

Ok dude. The square root of something is always positive. So root x^2 will equal a positive number equal in magnitude to x.absolute of x- positive number equal in magnitude to x. The end

 

[oh well]

wtf man? it is not true.... square root of x^2 = +- x , not absolute value of x , unless x is a distance, speed (not velocity) etc. (anything that cannot be negative)

 

[Capt Rifle]

square root of x^2 is equal to plus or minusx, absolute value of x is always positive. so i dunno

 

[Kurosaki]

No the square root by itself is positive

 

[oh well]

square root of x^2 is equal to x, absolute value of x is always positive. So when you square root x^2 it will always be positive

Not sure if srs

 

[oh well]

No the square root by itself is positive

Mate get ur facts right

 

[Kurosaki]

Mate get ur facts right

Wikipedia- the designation square root refers to the principal positive square root
But yes if not using that designation I no what u mean lol. Not dat stupid

 

[deswa1]

When you say root 3, by convention you are assumed to mean the positive root. That's why in the quadratic formula say, its plus/minus the squareroot, because if you didn't have the plus/minus, you would only be taking the positive value.

Back to the original question, the squareroot of x^2 therefore must be the positive x solution, not -x, so its |x|

 

[DamTameNaken]

by convention the square root is always positive

x^2 = 16
in this case x can equal 4 or -4

sqrt(16)=x
x can only equal to 4

you always take the positive or principle root when you're square rooting. It's weird and it only works by convention but that's how it is.

 

[Kurosaki]

Lol oh well I got my mates to back me up . lol Deswa start deleting old posts

 

[Riproot]

When you say root 3, by convention you are assumed to mean the positive root. That's why in the quadratic formula say, its plus/minus the squareroot, because if you didn't have the plus/minus, you would only be taking the positive value.

Back to the original question, the squareroot of x^2 therefore must be the positive x solution, not -x, so its |x|

it's not |x| it's x.

y = |x| is not always the same as y = x

is that right?

 

[Carrotsticks]

The answer y=|x| is correct and this can be observed by considering y=root(f(x)), where f(x) =x^2, in a geometric sense.

 

[Riproot]

The answer y=|x| is correct and this can be observed by considering y=root(f(x)), where f(x) =x^2, in a geometric sense.

 

[OH1995]

The answer y=|x| is correct and this can be observed by considering y=root(f(x)), where f(x) =x^2, in a geometric sense.

+1 haha
My maths teacher always drills into us that the square root of a perfect square is in fact the absolute value.

 

[Alext_]

bumpp.
i'm annoyed how i still can't grasp this concept...
can someone please explain in a more 'simpler' way...or 'complex' way if it's easier to explain that way

 

[Alext_]

'the squareroot of x^2 therefore must be the positive x solution, not -x, so its |x|' - Kurosaki
i don't really understand that last part: ' not -x, so its |x|'

 

[Trebla]

I'll try with an example.



Let's consider for the example, the case that x = 2





Hence in this case

Now consider the case that x = -2





Hence in this case

Now think about extensions to this for every single real number of x (if you can).

 

[Alext_]

thankyou so much. i finally understand it