Is sqrt((x-3)^2) = 3 - x?

[RonPurewal]

Rahul, as mentioned by Ron, the square root of a number is always positive in GMATLand.

correct.

not just gmatland -- this is actually a universal mathematical convention, i.e., something that is obeyed by everyone who uses the "√" sign anywhere.

[sukrutip]

I dont get it.

I see two eqautions.
x-3=3-x
2x=6
x=3

x-3=x-3
x=anything

How are you getting x <= 3 or x negative?

[tim]

those two equations are not universally true. each is only true under certain conditions. go back and pay careful attention to that part of the explanation..

[steven.sheph]

I dont get it.

I see two eqautions.
x-3=3-x
2x=6
x=3

x-3=x-3
x=anything

How are you getting x <= 3 or x negative?

I am with you. I get the exact same answer and am now confused by this.

[jlucero]

Anytime you have even exponents or absolute values, you need to solve for two conditions- one when the solution is positive and one when the solution is negative. Here's a simple example:

|x| = 5
A) If the number inside the absolute value is positive, then:
x = 5

B) If the number inside the absolute value is negative, then:
-(x) = 5
x = -5

You do this in your head without thinking, but that's the same method Ron explains in this problem. Think of squaring a number and then square rooting it as an absolute value (sqrt((-2)^2) = 2). So in this problem:

|x-3| = 3-x?
A) If the number inside the absolute value is positive, then:
(x-3) = 3-x
2x = 6
x=3 WHENEVER condition A stands

B) If the number inside the absolute value is negative, then:
-(x-3) = 3-x
3-x = 3-x
always true WHENEVER condition B stands
when does condition B stand?
whenever x-3 is negative
x-3 < 0
x < 3

So combining these two equations, we see x = 3 OR x < 3 means x <=3

[supratim7]

Would following takeaways be good?
Are there anymore caveats as far as radical sign, exponents & roots, and absolute value are concerned?

A) For an unknown term, e.g. x, √(x^2) = |x|
* x could be negative, 0, or positive.

B) For a known positive term, e.g. 9, √9 = 3, and for 0, √0 = 0

C) If x^2 = y^2, then x = ± y
* x, y could be negative, 0, or positive.

D) If |x| = y, then x = y when x ≥ 0 OR x = -y when x ≤ 0
* x, y could be negative, 0, or positive.

Also, what happens in Planet GMAT when the appearance changes from √(x^2) to (x^2)^1/2

Many thanks | Supratim

[jlucero]

Would following takeaways be good?
Are there anymore caveats as far as radical sign, exponents & roots, and absolute value are concerned?

A) For an unknown term, e.g. x, √(x^2) = |x|
* x could be negative, 0, or positive.

B) For a known positive term, e.g. 9, √9 = 3, and for 0, √0 = 0

C) If x^2 = y^2, then x = ± y
* x, y could be negative, 0, or positive.

D) If |x| = y, then x = y when x ≥ 0 OR x = -y when x ≤ 0
* x, y could be negative, 0, or positive.

Also, what happens in Planet GMAT when the appearance changes from √(x^2) to (x^2)^1/2

Many thanks | Supratim

All of this looks good except point D:

|x| = y

x could be negative, 0, or positive. However, since |x| must be positive, y must be a positive number.

And finally, yes, √(x^2) is the same as (x^2)^1/2

[supratim7]

Thank you for replying Joe. Appreciate it.

Are following iterations OK?

A) For an unknown term, e.g. x, √(x^2) = (x^2)^1/2 = |x|
* x could be negative, 0, or positive.

B) For a known positive term, e.g. 9, √9 = (9^2)^1/2 = 3, and for 0, √0 = (0^2)^1/2 = 0

D) If |x| = y, then x = y when x ≥ 0 OR x = -y when x ≤ 0
* x could be negative, 0, or positive.
* y MUST BE either 0 or positive

Many thanks | Supratim

[RonPurewal]

not sure why you're calling those "iterations""”but, yes, those are true.

(in math, "iterations" are consecutive repetitions of the same mathematical procedure or process. for instance, if 5% compound interest is added to a bank account each year, then that's a series of iterations.)

[supratim7]

I see.. I had a loose understanding of the word 'iteration'. "following modification/correction" is what I meant.

Thank you for highlighting. Appreciate it :)

[jnelson0612]

Good discussion--thanks all!

[asharma8080]

Ok - the simplification of square roots totally made me miss this problem. I had an error while simplifying this problem:

Simplifying question stem:
sqrt((x-3)^2) = 3 - x
or, square both sides:
(x-3)^2 = (3-x)^2
then, square root both sides:
x-3 = 3-x -- [u]I missed the step in red. Is this a matter of practicing the square roots equations?[/u]

Or more simply:
sqrt((x-3)^2) = 3 - x
square both sides
(x-3)^2 = (3-x)^2
sq root both sides
x-3 = 3-x

Plugging in:
statement 1
try x = 2
2-3 = 3-2
-1 not equal to 1.. NO
x = 4
4-3 = 4-3
1 = 1 .. Yes.

Statement 1 is NOT suff.

- x |x| > 0
Plugging in x = 2
-2|2| > 0 -- this won't hold true
Plugging in x = -2
-(-2) |-2| > 0 ... yes
Another value, x = -4
-(-4) |-4| > 0
So x < 0

Now, for negative numbers, plugging in:
x = -2
x-3 = 3-x
-2 - 3 = 3 +5
-5 is not equal to 5
similarly, x= -4
-4 - 3 = 3+4
-7 is not equal to 7.

Statement 2 is sufficient as the equation is NOT equal

[RonPurewal]

asharma8080, this thread is 40+ posts long, and this issue has already been treated -- quite comprehensively -- within the thread.

please read the entire thread. if you still have questions and/or things that you don't understand -- after you've read everything on the thread -- go ahead and post.

thanks.