- anonymous

ques

- katieb

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- anonymous

why is
sqrt{x^2}=|x|??
let
f(x)=sqrt{x^2}
g(x)=x
h(x)=|x|
at x= -1
we have
f(x)=plus minus 1
g(x)=-1
h(x)=1
clearly it looks like
sqrt{x^2}=plus minus x
sorry my equation box is giving errors right now so im using this

- ganeshie8

|dw:1435127988303:dw|

- anonymous

plus minus 2?

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## More answers

- ganeshie8

Also
|dw:1435128205480:dw|

- anonymous

|dw:1435128362002:dw|

- ganeshie8

we could say the "2nd roots of 4" are "+2 and -2" as both of them when rised to 2nd power give us 4; but if we stick to the definition of sqrt function, we only pick the positive root.

- ikram002p

@Nishant_Garg what is x^2 range ?

- ganeshie8

\(\sqrt{4} = 2\)

- anonymous

I'm confused, doesn't square rooting gives 2 values?

- ganeshie8

By definition, sqrt(x) is always the positive root of x

- ganeshie8

sqrt(x) is a single valued function

- anonymous

I don't understand, that sounds really really wrong

- ganeshie8

do you have definition of sqrt(x) function wid you ?

- ganeshie8

stick to that

- anonymous

Well, idk, I've been doing plus minus x all the time for sqrt of x^2
How do you suppose the equation
x^2=4
has 2 solutions then if u only take the positive value?

- ganeshie8

\[x^2=4 \implies x = \pm \sqrt{~4~} = \pm 2\]

- ganeshie8

\[x^2 =3 \implies x=\pm\sqrt{3} \]

- ganeshie8

i think you're confusing sqrt(x) function with the complex roots of x

- anonymous

This is melting my brain, it feels like I've been square rooting wrong the whole time...
so
is
x^2=4
different from
x=sqrt{4}
??

- ganeshie8

Yes they are not same.
\[x = \sqrt{4} \implies x^2=4\]
but the converse is not true.

- ganeshie8

\[x^2=4 \implies x=\color{red}{\pm}\sqrt{4}\]

- anonymous

and
x=sqrt{4}
implies
x=2 ONLY ?

- ganeshie8

there wont be any confusion if we take the definition seriosuly and stick to it

- ganeshie8

that is correct
\[x=\sqrt{4}\implies x=2\]

- anonymous

and not minus 2

- ganeshie8

Yep. sqrt(x) is never negative

- anonymous

This is just bizzare, I always thought sqrt gives plus minus.....It all looks like a different point of view now

- anonymous

then the equation
x^2=4
is to be solved like
sqrt{x^2}=sqrt{4}
implies |x|=2(by definition of sqrt)
removing the mod,
x=plus minus 2
but for the equation
x=sqrt{4}
we have
x=2(by definition of sqrt)

- ganeshie8

Haha ikr! one not so cool thing about mathematics is sometimes we just need to follow rules and stick to definitions even though they seem unintuitive

- anonymous

can u check my post

- ganeshie8

that looks good to me!

- anonymous

so the plus minus is a consequence of the modulus function and not square root

- ganeshie8

modulus function is a consequence of sqrt

- ganeshie8

|dw:1435129510602:dw|

- ganeshie8

as you said |x|=2 has two solutions

- anonymous

makes sense now but feels a little different :)

- ganeshie8

since we're messing with roots, notice that below is perfectly okay :
|dw:1435129597807:dw|

- ganeshie8

but below is wrong :
|dw:1435129708335:dw|

- anonymous

should be
|X^7|?

- ganeshie8

Yes! |x^7| or |x|^7
both are fine

- anonymous

as it has an odd power it can gives negative numbers

- ganeshie8

Exactly! which changes the original function sqrt(x^14) as x^14 is always positive

- anonymous

amazing thanks a lot...

- ganeshie8

np:)

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