Question

Why do we have to put an absolute value in the final answer?!

Answer 1

final answer to what?

Answer 2

\[\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}\]

Answer 3

Can't even see that, is it 10 th root?

Answer 4

Yup

Answer 5

\[\sqrt[10]{x^10}\]

Answer 6

see moderator has warned me now...whats the way to delet ur account...can somebody give the link...i m going frm here

Answer 7

What? O.o

Answer 8

can u pls give me the link...i want to delet my account..i can't find it anywhere

Answer 9

I don't know.. You have to contact a moderator

Answer 10

cmon moderator show up now...

Answer 11

The only reason I can think of is they want to make sure you get a real and postiive (as opposed to complex) root. There will be 10 different roots.

Answer 12

@akash_809 you cannot delete OS accounts

Answer 13

But you can write it without absolute value, right? Because (or this is what I've been taught) there's x^3 in the radicand so it's definitely positive..

Answer 14

the simple reason is that\[x^{10}=(-x)^{10}\]so \[\large\sqrt[10]{x^{10}}\]will give you the same answer whether x is negative or positive

Answer 15

what...thats cynical...open study community is surely going nowher with all these warnings and suspensions...and to top that off , you even cant delete your account...what is this open study team please response...i wnat to know how can i delete my account

Answer 16

absolute value of x is actually defined in 2 ways:
1) distance from the point x=0 on the number line
2) \(|x|=\sqrt{x^2}\)

Answer 17

Thanks :))

Answer 18

so since\[\sqrt{(-x)^2}=\sqrt{x^2}=|x|\]then\[\sqrt[10]{x^{10}}=\sqrt[10]{(-x)^{10}}=|x|\](because x to any even power is positive, whether x>0 or x<0)

Answer 19

does that help you @aroub ?

Answer 20

welcome :D

Answer 21

In this case I totally understand why. But in the question itself there's an odd exponent in the radicand so it's definitely positive right? So then no need for an absolute value..

Answer 22

@TuringTest root of negative number is not defined...it will result in iota value...so -x wont be the answer i think

Answer 23

not necessarily, let's assume x<0...

Answer 24

If you assumed that, the question will be wrong.

Answer 25

-I think-

Answer 26

yes it is...\[lets take an example......\sqrt{-2} is \not defined unless you are taking imaginary numbers into account then u wl get +-1.414 i\]

Answer 27

the root of a negative number is not undefined, it is imaginary...\[\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}=\sqrt[10]{x^4}\times\sqrt[10]{x^3\cdot x^3}\]

Answer 28

sry reply got messed up i was just saying that square root of -2 is not defined unless you consider imaginary numbers as wel..

Answer 29

Put imaginary numbers on a side for now..

Answer 30

imaginary numbers are defined...
now if x<0, then x^3<0
that means that (x^3)(x^3)>0
which is what we would expect given that x^6>0 since all numbers to a positive exponent are >0

Answer 31

another way to see it, let's assume x>0\[\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}=\sqrt[10]{x^4}\times\sqrt[10]{x^3\cdot x^3}=\sqrt[10]{x^{10}}=|x|\]now we do it with the negative of x\[\sqrt[10]{(-x)^4} \times \sqrt[10]{(-x)^3} \times \sqrt[10]{(-x)^3}=\sqrt[10]{(-x)^4}\times\sqrt[10]{(-x)^3\cdot (-x)^3}\]\[=\sqrt[10]{(-x)^{10}}=\sqrt[10]{x^{10}}=|x|\]so you see, after all is said and done we do not know whether the x we had at the beginning was positive or negative, hence we can only conclude that the answer is |x|

Answer 32

Okay take this as another example:

\[4\sqrt[4]{5a^2}\times 8\sqrt[4]{125a}\times 2\sqrt[4]{a}\]

Here, a does not need an absolute value. Why? You can assume that a>0

Answer 33

yes you would have to assume a >0

Answer 34

let's just look at powers of a...

Answer 35

if a>0\[\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=|a|=a\](since we already knew a>0
now with the negative of a...

Answer 36

yea i m waiting for that negative values of a...what would it give

Answer 37

Aoh, so you have to try the negative and the positive?

Answer 38

\[\sqrt[4]{(-a)^2}\times\sqrt[4]{-a}\times\sqrt[4]{-a}=\sqrt[4]{a^4}=|a|\neq a\]because |a|>0, but we just assumed a<0 !
a contradiction, hence the assumption makes a difference

Answer 39

sorry, I mean we took the negative of a...

Answer 40

so \[|a|\neq a\]

Answer 41

let me write that again, gotta be careful here

Answer 42

that's the problem...the thing you did here is called fallacy of maths... http://en.wikipedia.org/wiki/Mathematical_fallacy ....search invalid roots in this article

Answer 43

assume a>0\[\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=|a|=a\]assume a<0, then\[\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=|a|\neq a\]since we assumed a<0

Answer 44

@akash_809 yes but at least one of these numbers is positive because it is raised to an even power, so I have committed no fallacy

Answer 45

did u went through the link......the proofs given like \[1=\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}=i.i=-1 ....so u hv 1=-1...and certainly \it is \not valid..\]

Answer 46

same thing with the earlier problem...

Answer 47

and 1=-1....surely not right...

Answer 48

yes, but\[\sqrt{(-1)^2}\cdot\sqrt{-1}=\sqrt1\cdot\sqrt{-1}\] is not the same as\[\sqrt{-1}\cdot\sqrt{-1}\]

Answer 49

notice we have a^2 in the above, so we know *that* part is positive

Answer 50

this turned out to be more subtle than I bet you realized @aroub :)

Answer 51

lol, yeah true xD ALWAYS thought absolute values are so easy! And whenever someone asks me about absolute values I answer until this question came up -.-' This question is a bit confusing >.< Well, THANKS LOOAADSSSSSS :D

Answer 52

welcome :)