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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: kenhung123 on March 13, 2010, 06:15:15 pm

Title: Is this simplifyable?
Post by: kenhung123 on March 13, 2010, 06:15:15 pm: $f(x)=\frac{1}{\sqrt{x^2}}=\frac{1}{x}$
As $\sqrt{x^2}=x$
Title: Re: Is this simplifyable?
Post by: TrueTears on March 13, 2010, 06:16:39 pm: $\sqrt{x^2} = |x|$ unless restrictions are applied.
Title: Re: Is this simplifyable?
Post by: kenhung123 on March 13, 2010, 07:15:20 pm: But I thought the root of x^2 is the positive only?
Title: Re: Is this simplifyable?
Post by: physics on March 13, 2010, 07:20:05 pm: i think its simplifyable .... i think u should try it on the CAS calculator to see if it simplifys for you or type that equation as equalling the one u think it would and if it says true then it is ..if it says false well then its false >< i ope u get what i mean
Title: Re: Is this simplifyable?
Post by: kyzoo on March 13, 2010, 07:24:58 pm: Quote from: kenhung123 on March 13, 2010, 06:15:15 pm
As $\sqrt{x^2}=x$

>.< you would have lost tons of marks on last year's Methods exams if you did that
Title: Re: Is this simplifyable?
Post by: kenhung123 on March 13, 2010, 07:46:51 pm: ^ ???
Title: Re: Is this simplifyable?
Post by: superflya on March 13, 2010, 07:48:34 pm: huh

thered be two cases wouldnt there?
Title: Re: Is this simplifyable?
Post by: stonecold on March 13, 2010, 07:50:13 pm: Quote from: kyzoo on March 13, 2010, 07:24:58 pm
Quote from: kenhung123 on March 13, 2010, 06:15:15 pm
As $\sqrt{x^2}=x$

>.< you would have lost tons of marks on last year's Methods exams if you did that

true!

modulus x does not just equal x :)
Title: Re: Is this simplifyable?
Post by: TrueTears on March 13, 2010, 07:51:08 pm: Quote from: kenhung123 on March 13, 2010, 07:15:20 pm
But I thought the root of x^2 is the positive only?
u sure?
Title: Re: Is this simplifyable?
Post by: kenhung123 on March 13, 2010, 07:53:12 pm: According to heinemann..
Title: Re: Is this simplifyable?
Post by: TrueTears on March 13, 2010, 07:54:02 pm: Also I'd like to put this question out there (where is the fallacy?)

$+\sqrt{(-3)^2} = +\sqrt{9} = +3$

But

$+\sqrt{(-3)^2} = + (-3^2)^{\frac{1}{2}} = + (-3)^{\frac{2}{2}} = -3$

Thus $+3 = -3$

[Think about multivalued roots!]
Title: Re: Is this simplifyable?
Post by: simonhu81292 on March 13, 2010, 07:55:40 pm: i think it it's saying when only applied to certain situations in applications .. such that you can't have a neg answer...
Title: Re: Is this simplifyable?
Post by: TrueTears on March 13, 2010, 07:56:33 pm: Oh and I forgot to say the fallacy is beyond methods, don't worry if you can't understand why, it is not very important in VCE maths :P
Title: Re: Is this simplifyable?
Post by: kenhung123 on March 13, 2010, 08:00:45 pm: Quote from: simonhu81292 on March 13, 2010, 07:55:40 pm
i think it it's saying when only applied to certain situations in applications .. such that you can't have a neg answer...
But....how do they expect us to assume that...
Title: Re: Is this simplifyable?
Post by: the.watchman on March 13, 2010, 08:40:17 pm: Yeah, $\sqrt{x^2} = |x|$

BUT you can have $(\sqrt{x})^2 = x$
Title: Re: Is this simplifyable?
Post by: kyzoo on March 13, 2010, 09:02:06 pm: Quote from: kenhung123 on March 13, 2010, 07:46:51 pm
^ ???

http://www.vcaa.vic.edu.au/vcaa/vce/studies/mathematics/methods/assessreports/2009/mm2_assessrep_09.pdf
Methods 2009 Exam 2 Assessor's report

Question 1 e i.)
Quote
Many students showed a poor understanding of the absolute value function and incorrectly simplified √(x²) to x.
Title: Re: Is this simplifyable?
Post by: kenhung123 on March 13, 2010, 09:04:47 pm: I don't know man, that's what the textbook says
Title: Re: Is this simplifyable?
Post by: the.watchman on March 13, 2010, 09:12:09 pm: Yeah, make sure you can tell the difference between $\sqrt{x^2}$ AND $(\sqrt{x})^2$
Title: Re: Is this simplifyable?
Post by: kenhung123 on March 13, 2010, 09:13:02 pm: The text looks like the first one..
Title: Re: Is this simplifyable?
Post by: superflya on March 13, 2010, 09:13:51 pm: Quote from: kenhung123 on March 13, 2010, 09:13:02 pm
The text looks like the first one..

so u cant simplify it to $x$
Title: Re: Is this simplifyable?
Post by: kenhung123 on March 13, 2010, 09:23:53 pm: It says it =x not +/-x
Title: Re: Is this simplifyable?
Post by: the.watchman on March 13, 2010, 09:25:32 pm: What, it says $\sqrt{x^2} = x$ ?

Well then it's wrong :)
Title: Re: Is this simplifyable?
Post by: kenhung123 on March 13, 2010, 09:26:22 pm: Didn't you see the attachment?
Title: Re: Is this simplifyable?
Post by: TrueTears on March 13, 2010, 09:29:21 pm: books are written by humans

humans are not perfect

but practise makes perfect

but nobody is perfect

so why practise?
Title: Re: Is this simplifyable?
Post by: simonhu81292 on March 13, 2010, 09:42:06 pm: wow...pro~~~
i liked that~~~
0 motivation now..
no pressure ~~ !!
haha
Title: Re: Is this simplifyable?
Post by: the.watchman on March 13, 2010, 09:43:23 pm: THIS IS A FAIL TEXTBOOK FOR YOUR INFORMATION!!! :D

If x=-3, $\sqrt{(-3)^2} = 3 = -x$
Title: Re: Is this simplifyable?
Post by: TrueTears on March 13, 2010, 09:44:44 pm: Quote from: the.watchman on March 13, 2010, 09:43:23 pm
THIS IS A FAIL TEXTBOOK FOR YOUR INFORMATION!!! :D

If x=-3, $\sqrt{(-3)^2} = 3 = -x$
no...

$\sqrt{(-3)^2} = (-3)^{\frac{2}{2}} = -3... = x...$
Title: Re: Is this simplifyable?
Post by: physics on March 13, 2010, 09:46:23 pm: Quote from: TrueTears on March 13, 2010, 09:44:44 pm
Quote from: the.watchman on March 13, 2010, 09:43:23 pm
THIS IS A FAIL TEXTBOOK FOR YOUR INFORMATION!!! :D

If x=-3, $\sqrt{(-3)^2} = 3 = -x$
no...

$\sqrt{(-3)^2} = (-3)^{\frac{2}{2}} = -3... = x...$
there was this sorta question on last years exam 1 and 2 i think...no wonder i stuffed up exam 1 XD
Title: Re: Is this simplifyable?
Post by: the.watchman on March 13, 2010, 09:47:33 pm: Quote from: TrueTears on March 13, 2010, 09:44:44 pm
Quote from: the.watchman on March 13, 2010, 09:43:23 pm
THIS IS A FAIL TEXTBOOK FOR YOUR INFORMATION!!! :D

If x=-3, $\sqrt{(-3)^2} = 3 = -x$
no...

$\sqrt{(-3)^2} = (-3)^{\frac{2}{2}} = -3... = x...$

Erm ... isn't it $\sqrt{(-3)^2} = \sqrt{9} = 3 = -x$ (because of +ve square root?)
Title: Re: Is this simplifyable?
Post by: TrueTears on March 13, 2010, 09:47:44 pm: the reason in methods $\sqrt{x^2} = |x|$ is because the square roots are multivalued, thus the modulus get's rid of this misconception. You will learn more about it in uni.
Title: Re: Is this simplifyable?
Post by: TrueTears on March 13, 2010, 09:48:32 pm: Quote from: the.watchman on March 13, 2010, 09:47:33 pm
Quote from: TrueTears on March 13, 2010, 09:44:44 pm
Quote from: the.watchman on March 13, 2010, 09:43:23 pm
THIS IS A FAIL TEXTBOOK FOR YOUR INFORMATION!!! :D

If x=-3, $\sqrt{(-3)^2} = 3 = -x$
no...

$\sqrt{(-3)^2} = (-3)^{\frac{2}{2}} = -3... = x...$

Erm ... isn't it $\sqrt{(-3)^2} = \sqrt{9} = 3 = -x$ (because of +ve square root?)
not rly...

where would the fallacy be in my step then?
Title: Re: Is this simplifyable?
Post by: physics on March 13, 2010, 09:49:05 pm: Quote from: TrueTears on March 13, 2010, 09:47:44 pm
the reason in methods $\sqrt{x^2} = |x|$ is because the square roots are multivalued, thus the modulus get's rid of this misconception. You will learn more about it in uni.
we learn this in uni maths?...i dont look foward to it...:(
Title: Re: Is this simplifyable?
Post by: the.watchman on March 13, 2010, 09:52:17 pm: Quote from: TrueTears on March 13, 2010, 09:48:32 pm
Quote from: the.watchman on March 13, 2010, 09:47:33 pm
Quote from: TrueTears on March 13, 2010, 09:44:44 pm
Quote from: the.watchman on March 13, 2010, 09:43:23 pm
THIS IS A FAIL TEXTBOOK FOR YOUR INFORMATION!!! :D

If x=-3, $\sqrt{(-3)^2} = 3 = -x$
no...

$\sqrt{(-3)^2} = (-3)^{\frac{2}{2}} = -3... = x...$

Erm ... isn't it $\sqrt{(-3)^2} = \sqrt{9} = 3 = -x$ (because of +ve square root?)
not rly...

where would the fallacy be in my step then?

No idea >.<
Title: Re: Is this simplifyable?
Post by: TrueTears on March 13, 2010, 09:52:48 pm: there isn't and both are right :)
Title: Re: Is this simplifyable?
Post by: kenhung123 on March 13, 2010, 09:54:32 pm: Quote from: TrueTears on March 13, 2010, 09:48:32 pm
Quote from: the.watchman on March 13, 2010, 09:47:33 pm
Quote from: TrueTears on March 13, 2010, 09:44:44 pm
Quote from: the.watchman on March 13, 2010, 09:43:23 pm
THIS IS A FAIL TEXTBOOK FOR YOUR INFORMATION!!! :D

If x=-3, $\sqrt{(-3)^2} = 3 = -x$
no...

$\sqrt{(-3)^2} = (-3)^{\frac{2}{2}} = -3... = x...$

Erm ... isn't it $\sqrt{(-3)^2} = \sqrt{9} = 3 = -x$ (because of +ve square root?)
not rly...

where would the fallacy be in my step then?

Maybe need to follow BEDMAS
Title: Re: Is this simplifyable?
Post by: TrueTears on March 13, 2010, 09:57:50 pm: ok.
Title: Re: Is this simplifyable?
Post by: the.watchman on March 13, 2010, 09:58:20 pm: Quote from: TrueTears on March 13, 2010, 09:52:48 pm
there isn't and both are right :)

LOL :D
Call it a draw this time ;)

(actually, is it ok to skip from $\sqrt{(-3)^2}$ to $(-3)^\frac{2}{2}$ ?
it should be right, but that seems to be where we diverge :P)
Title: Re: Is this simplifyable?
Post by: TrueTears on March 13, 2010, 10:04:26 pm: i didn't skip anything
Title: Re: Is this simplifyable?
Post by: /0 on March 13, 2010, 10:11:01 pm: isn't $\sqrt{}$ defined as the positive root?
Title: Re: Is this simplifyable?
Post by: the.watchman on March 13, 2010, 10:11:42 pm: Quote from: /0 on March 13, 2010, 10:11:01 pm
isn't $\sqrt{}$ defined as the positive root?

Yeah, so the answer should be positive...
Oh whatever! :P
Title: Re: Is this simplifyable?
Post by: TrueTears on March 13, 2010, 10:12:53 pm: Quote from: /0 on March 13, 2010, 10:11:01 pm
isn't $\sqrt{}$ defined as the positive root?
definitions?
Title: Re: Is this simplifyable?
Post by: /0 on March 13, 2010, 10:13:40 pm: Quote from: TrueTears on March 13, 2010, 10:12:53 pm
Quote from: /0 on March 13, 2010, 10:11:01 pm
isn't $\sqrt{}$ defined as the positive root?
definitions?

"Every non-negative real number x has a unique non-negative square root,
called the principal square root, which is denoted with a radical sign as $\sqrt{x}$ "

- Wikipedia
Title: Re: Is this simplifyable?
Post by: TrueTears on March 13, 2010, 10:14:23 pm: i wonder what kamil would say about this
Title: Re: Is this simplifyable?
Post by: kamil9876 on March 14, 2010, 11:27:01 pm: quite frankly I don't really care. I'm more worried about the pedantic and elitist image of 'uni maths' that has probably scared future generations and may possibly cause loss of some great talent.

Quote
we learn this in uni maths?...i dont look foward to it... :(
Title: Re: Is this simplifyable?
Post by: Mao on March 15, 2010, 12:22:17 am: I agree with /0, the case that $\sqrt{x^2} = |x|$ is not because the square roots are multivalued. If that was the case, it would be expressed as $\sqrt{x^2} = \pm x$ [see how it's multivalued?] HOWEVER, this is not true.

The square root function is defined as the positive branch, the fact that it returns the absolute value is because $x^2$ is not one-to-one, and whenever a negative value is substituted, a positive value is returned (hence, absolute).

Quote from: kenhung123 on March 13, 2010, 07:15:20 pm
But I thought the root of x^2 is the positive only?
^^ that statement is correct, hence you cannot say $\sqrt{x^2} = x$ , because this means the root could be negative (for $x \le 0$ )
Title: Re: Is this simplifyable?
Post by: the.watchman on March 15, 2010, 06:19:45 am: Quote from: Mao on March 15, 2010, 12:22:17 am
I agree with /0, the case that $\sqrt{x^2} = |x|$ is not because the square roots are multivalued. If that was the case, it would be expressed as $\sqrt{x^2} = \pm x$ [see how it's multivalued?] HOWEVER, this is not true.

The square root function is defined as the positive branch, the fact that it returns the absolute value is because $x^2$ is not one-to-one, and whenever a negative value is substituted, a positive value is returned (hence, absolute).

Quote from: kenhung123 on March 13, 2010, 07:15:20 pm
But I thought the root of x^2 is the positive only?
^^ that statement is correct, hence you cannot say $\sqrt{x^2} = x$ , because this means the root could be negative (for $x \le 0$ )

YES thank you for confirmation :)
Title: Re: Is this simplifyable?
Post by: TrueTears on March 16, 2010, 08:51:34 am: Quote from: kamil9876 on March 14, 2010, 11:27:01 pm
quite frankly I don't really care. I'm more worried about the pedantic and elitist image of 'uni maths' that has probably scared future generations and may possibly cause loss of some great talent.

Quote
we learn this in uni maths?...i dont look foward to it... :(
exactly, definitions are just definitions.