anonymous
  • anonymous
ques
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • ganeshie8
|dw:1435127988303:dw|
anonymous
  • anonymous
plus minus 2?

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ganeshie8
  • ganeshie8
Also |dw:1435128205480:dw|
anonymous
  • anonymous
|dw:1435128362002:dw|
ganeshie8
  • 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
  • ikram002p
@Nishant_Garg what is x^2 range ?
ganeshie8
  • ganeshie8
\(\sqrt{4} = 2\)
anonymous
  • anonymous
I'm confused, doesn't square rooting gives 2 values?
ganeshie8
  • ganeshie8
By definition, sqrt(x) is always the positive root of x
ganeshie8
  • ganeshie8
sqrt(x) is a single valued function
anonymous
  • anonymous
I don't understand, that sounds really really wrong
ganeshie8
  • ganeshie8
do you have definition of sqrt(x) function wid you ?
ganeshie8
  • ganeshie8
stick to that
anonymous
  • 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
  • ganeshie8
\[x^2=4 \implies x = \pm \sqrt{~4~} = \pm 2\]
ganeshie8
  • ganeshie8
\[x^2 =3 \implies x=\pm\sqrt{3} \]
ganeshie8
  • ganeshie8
i think you're confusing sqrt(x) function with the complex roots of x
anonymous
  • 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
  • ganeshie8
Yes they are not same. \[x = \sqrt{4} \implies x^2=4\] but the converse is not true.
ganeshie8
  • ganeshie8
\[x^2=4 \implies x=\color{red}{\pm}\sqrt{4}\]
anonymous
  • anonymous
and x=sqrt{4} implies x=2 ONLY ?
ganeshie8
  • ganeshie8
there wont be any confusion if we take the definition seriosuly and stick to it
ganeshie8
  • ganeshie8
that is correct \[x=\sqrt{4}\implies x=2\]
anonymous
  • anonymous
and not minus 2
ganeshie8
  • ganeshie8
Yep. sqrt(x) is never negative
anonymous
  • anonymous
This is just bizzare, I always thought sqrt gives plus minus.....It all looks like a different point of view now
anonymous
  • 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
  • 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
  • anonymous
can u check my post
ganeshie8
  • ganeshie8
that looks good to me!
anonymous
  • anonymous
so the plus minus is a consequence of the modulus function and not square root
ganeshie8
  • ganeshie8
modulus function is a consequence of sqrt
ganeshie8
  • ganeshie8
|dw:1435129510602:dw|
ganeshie8
  • ganeshie8
as you said |x|=2 has two solutions
anonymous
  • anonymous
makes sense now but feels a little different :)
ganeshie8
  • ganeshie8
since we're messing with roots, notice that below is perfectly okay : |dw:1435129597807:dw|
ganeshie8
  • ganeshie8
but below is wrong : |dw:1435129708335:dw|
anonymous
  • anonymous
should be |X^7|?
ganeshie8
  • ganeshie8
Yes! |x^7| or |x|^7 both are fine
anonymous
  • anonymous
as it has an odd power it can gives negative numbers
ganeshie8
  • ganeshie8
Exactly! which changes the original function sqrt(x^14) as x^14 is always positive
anonymous
  • anonymous
amazing thanks a lot...
ganeshie8
  • ganeshie8
np:)

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