Math help

[dukemaster4]

He newgrounds I need help with this math problem.

are the functions f(x) = |x| and g(x) = (square root)x(squared) equal?

[fawnret]

why not ask a math site. here we will just flame you

[X-TERRORIST-X]

newgrounds isnt the place to ask for math help.. you'll probaly just get 20 'do a barrel rolls'

do a barrel roll

[Serbian-terrorist]

No, g(x) can be both positive and negative, while f(x) will always be only positive.

[dukemaster4]

At 10/3/07 09:57 PM, X-TERRORIST-X wrote: newgrounds isnt the place to ask for math help.. you'll probaly just get 20 'do a barrel rolls'

do a barrel roll

Since when did NG turn into /b/?

[americanidiot2f]

Don't ask NG homework help!

[jarenfox2004]

the square root of x squared is x so thats like saying does lxl = x. no because what if x was negative

[amaterasu]

At 10/3/07 09:56 PM, dukemaster4 wrote: He newgrounds I need help with this math problem.

are the functions f(x) = |x| and g(x) = (square root)x(squared) equal?

yes

[reality-check7]

At 10/3/07 09:56 PM, dukemaster4 wrote: He newgrounds I need help with this math problem.

are the functions f(x) = |x| and g(x) = (square root)x(squared) equal?

Yes, they are.

x = -2

f(-2) = |-2| = 2
g(-2) = sqrt( (-2)^2 ) = sqrt(4) = 2

f(x) = g(x)

[amaterasu]

At 10/3/07 09:58 PM, Serbian-terrorist wrote: No, g(x) can be both positive and negative, while f(x) will always be only positive.

no, because anything + or - squared results in a positive number, then square root that and you get x.

I can do the proof but I forgot one step involving something squared and the absolute value.

[Krisddd]

At 10/3/07 10:36 PM, reality-check7 wrote:

At 10/3/07 09:56 PM, dukemaster4 wrote: He newgrounds I need help with this math problem.

are the functions f(x) = |x| and g(x) = (square root)x(squared) equal?

Yes, they are.

x = -2

f(-2) = |-2| = 2
g(-2) = sqrt( (-2)^2 ) = sqrt(4) = 2

f(x) = g(x)

wouldnt the sqrt if x^2 just be x? y would u sqaure it first?

[LockDown68]

They aren't equal just take that at face value.

[NightmareFire]

Okay, that just made my brain hurt.

[Warrickneff]

At 10/3/07 10:54 PM, NightmareFire wrote: Okay, that just made my brain hurt.

Just think about it this way. Anything squared is positive. Anything in absolute value is positive. Since you immediatly take the square root of the squared function, it will return to it's original value.

The answer posted above is correct as well.

[reality-check7]

At 10/3/07 10:48 PM, Krisddd wrote:

At 10/3/07 10:36 PM, reality-check7 wrote:

At 10/3/07 09:56 PM, dukemaster4 wrote: He newgrounds I need help with this math problem.

are the functions f(x) = |x| and g(x) = (square root)x(squared) equal?

Yes, they are.

x = -2

f(-2) = |-2| = 2
g(-2) = sqrt( (-2)^2 ) = sqrt(4) = 2

f(x) = g(x)

wouldnt the sqrt if x^2 just be x?

Only if x is positive.

y would u sqaure it first?

...Because that's how math works. Imagine evaluating 2*(1+4). You wouldn't multiply 2 and 1, and then add 4. You'ld add 1 and 4 and then multiply by two. It's the same thing with square roots.

[bigjonny13]

|x| means that it is the relative value
so it is always a postive number
if its |-2| it is 2
if its |2| it is also 2

but when u have (root)2(squared) [same if it was -2]
=(root)4
= -2 or +2

so therefore f(x) not equal g(x)

[reality-check7]

At 10/3/07 11:13 PM, big-jonny-13 wrote: but when u have (root)2(squared) [same if it was -2]
=(root)4
= -2 or +2

True, but the principal square root is always positive. And when people refer to the square root function, they are usually reffering to the principal square root function.

So f(x) would equal g(x).

[EndGameOmega]

Well this might be a bit late but no they aren't in general equal. Many of you have been taut that the absolute value is simply sqrt(x^2) = |x|, but this isn't completely correct. Only in the case that x is real is this true. The absolute value is defined as |x| = sqrt(x*x), where *x is the complex conjugate of x. In the case that Im(x) = 0 then the complex conjugate is identical to x^2, i.e. x*x=xx=x^2, however if Im(x) = b =/=0 then the absolute value is:
sqrt(x*x)
sqrt[(a+ib)(a-ib)]
sqrt[a^2 + iab - iab - (i)^2b]
sqrt[a^2 + b^2],

where as:
sqrt(x^2)
sqrt[(a + ib)(a + ib)]
sqrt[( a^2 + iab + iab + (i)^2b^2)]
sqrt(a^2 + 2iab - b^2)

Obviously they are not in general equal. Hope that helps.

Note, i=sqrt(-1).