Absolute Value Confusion

Question

I was wondering if anyone could help me with the theory behind why sqrt(x^2)/x = |x|/x?
Intuitively it looks like the square root and exponent would cancel out, leaving the value to be 1. So in this situation, would you always assume that you'd square x, and then square root that value first before simplifying? (in that case, I can see why you would have the absolute value)
Thank you!

vvbalaji · Answer

I think you already got why sqrt(x^2) is same as |x|. In general while dealing with exponents of the form x/y we could apply the x first and then the root y or vice versa or even simplify x/y. The exact procedure to follow is a function of the actual problem.

stuckatfailure · Answer

Sorry I don't quite understand what you mean by "apply the x first"
do you mean I should run the operations on the X? From the answer, it seems that canceling operations does not take precedence over actual operations....

walker · Answer

Maybe I miss something but here are my thoughts:

1)  sqrt{x^2}  = |x|. Both operations just eliminate a negative sign for x. For example,  sqrt{(-1)^2}  = 1

2) |x|/x is 1 for positive, -1 for negative and undefined for 0. In other words, it is a way to get x's sign if x is non-zero.

stuckatfailure · Answer

Hi Walker,

I see what you mean. My question is exactly your first point:  sqrt{(-1)^2}  = 1. When I saw the question, I felt that  sqrt{(-1)^2}  = -1 since the square and square root would just cancel out. I'm assuming that's wrong? Is that a matter of order of operations? or is it just an identity we have to know?

walker · Answer

I see your point. You are right for any non-negative numbers but for negative

\((x^2)^{\frac12} \ne x^1\)

By the way, take a look at https://en.wikipedia.org/wiki/Exponentiation
They discuss the problem with negative base.

Bunuel · Answer

The point here is that  square root function can not give negative result :  \sqrt{expression}\geq{0} .

So  \sqrt{x^2}\geq{0} . But what does  \sqrt{x^2}  equal to?

Let's consider following examples:
If  x=5  -->  \sqrt{x^2}=\sqrt{25}=5=x=positive ;
If  x=-5  -->  \sqrt{x^2}=\sqrt{25}=5=-x=positive .

So we got that:
 \sqrt{x^2}=x , if  x\geq{0} ; 
 \sqrt{x^2}=-x , if  x<0 .

What function does exactly the same thing? The absolute value function! That is why  \sqrt{x^2}=|x|

More on this issue: 
GMAT is dealing only with   Real Numbers  : Integers, Fractions and Irrational Numbers.

Any nonnegative real number has a  unique non-negative square root  called  the principal square root  and unless otherwise specified,  the square root  is generally taken to mean  the principal square root .

When the GMAT provides the square root sign for an even root, such as  \sqrt{x}  or  \sqrt {x} , then the  only accepted answer is the positive root .

That is,  \sqrt{25}=5 , NOT +5 or -5. In contrast, the equation  x^2=25  has TWO solutions, +5 and -5.  Even roots have only non-negative value on the GMAT.

Odd roots will have the same sign as the base of the root. For example,  \sqrt {125} =5  and  \sqrt {-64} =-4 .

Other notes on this issue: 
 \sqrt {positive}=positive :  \sqrt{25}=5 . Even roots have only a positive value on the GMAT.

\sqrt {negative}=undefined :  \sqrt{-25}=undefined . Even roots from negative number is undefined on the GMAT (as GMAT is dealing only with   Real Numbers  ).

\sqrt {positive}=positive  and  \sqrt {negative}=negative :  \sqrt {125} =5  and  \sqrt {-64} =-4 . Odd roots will have the same sign as the base of the root.

You can also check Number Theory chapter of Math Book for more (link in my signature).

Hope it helps.

stuckatfailure · Answer

Thank you both for the help!!
Kudos!

Absolute Value Confusion

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