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by das.ashmita » Sat Oct 06, 2012 5:05 am
is root[(x-3)^2] = 3-x ?

1. x not equal to 3
2. -x|x| > 0

OA B[/list]

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by Brent@GMATPrepNow » Sat Oct 06, 2012 7:05 am
das.ashmita wrote:Is root[(x-3)^2] = 3-x ?

1. x not equal to 3
2. -x|x| > 0
Target question: Is root[(x-3)^2] = 3-x ?

This question is a great candidate for rephrasing the target question.

To begin, notice that we have two nice rules:
- If k > 0, then root(k^2) = k
- If k < 0, then root(k^2) = -k


Now observe that (3-x) = -(x-3)

Given the above information, under what conditions will root[(x-3)^2] = 3-x?
In other words, under what conditions will root[(x-3)^2] = -(x-3)?
This will occur if x-3 is negative.

So, we can now rephrase the target question as: Is (x-3) negative?
Or we can write: Is x-3 < 0?

. . . or better yet: Is x < 3?

Now that we've rephrased the target question in much simpler terms, we can check the statements.

Statement 1: x not equal to 3
This doesn't give us a definitive answer to the rephrased target question (Is x < 3? )
As such, statement 1 is NOT SUFFICIENT

Statement 2: -x|x| > 0
First notice that this implies that x does not equal zero.
Next, notice that, if x does not equal zero, then |x| will always be positive.
So, -x|x| > 0 is the same as saying (-x)(positive) > 0
In other words, the product (-x)(positive) results in a positive number.
This tells us that (-x) must be positive, which means x must be negative.
If x is negative, then x is definitely less than 3.
As such, statement 2 is SUFFICIENT

Answer = B

Cheers,
Brent
Last edited by Brent@GMATPrepNow on Sat Oct 06, 2012 7:14 am, edited 2 times in total.
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by Brent@GMATPrepNow » Sat Oct 06, 2012 7:10 am
If anyone is interested, we have a free video on the importance of rephrasing the target question: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100

Cheers,
Brent
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by das.ashmita » Sat Oct 06, 2012 8:39 am
Thanks a lot for the elaborate explanation :)

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by sachin_yadav » Fri Apr 12, 2013 8:47 am
After solving root[(x-3)^2] = 3-x, i got |x - 3| = 3 - x.
Now i can easily plug in few numbers according to the statement (1) & (2), and get the answer.

For example, (1) x not equal to 3.
I can use x = 0 and that satisfies the condition, but if i use x = 4, then definitely no.

Statement (2) simply gives that x < 0. So, Answer is B

Well this wasn't the first attempt that got me close to the answer. I wasted a lot of my time by solving absolute value equation further |x - 3| = 3 - x

(a). x - 3 = 3 - x
x + x = 3 + 3
2x = 6
x = 3

(b). x - 3 = -(3 - x)
x - 3 = -3 + x
x - x = -3 + 3
0 = 0

After solving all this i came to the conclusion that rephrase question is is x = 3 ?
Isn't this correct ?
Am i oversimplifying the equation ?

Please let me know your thoughts about this work.

Sachin
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by Anju@Gurome » Fri Apr 12, 2013 9:07 am
sachin_yadav wrote:I wasted a lot of my time by solving absolute value equation further |x - 3| = 3 - x

(a). x - 3 = 3 - x [This is possible if (x - 3) ≥ 0]
x + x = 3 + 3
2x = 6
x = 3 [This means only one value of x satisfies the equation in the range (x - 3) ≥ 0]

(b). x - 3 = -(3 - x) [This is possible if (x - 3) ≤ 0]
x - 3 = -3 + x
x - x = -3 + 3
0 = 0 [This means all the value of x in the range (x - 3) ≤ 0 satisfy the equation ]
Hence, the rephrased question should be is x ≤ 3?

While solving absolute value problems do not forget the definition of absolute value.

Hope that helps.
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by sachin_yadav » Sun Apr 14, 2013 2:14 am
Thank you so much and really appreciate it :D

Sachin
Anju@Gurome wrote:
sachin_yadav wrote:I wasted a lot of my time by solving absolute value equation further |x - 3| = 3 - x

(a). x - 3 = 3 - x [This is possible if (x - 3) ≥ 0]
x + x = 3 + 3
2x = 6
x = 3 [This means only one value of x satisfies the equation in the range (x - 3) ≥ 0]

(b). x - 3 = -(3 - x) [This is possible if (x - 3) ≤ 0]
x - 3 = -3 + x
x - x = -3 + 3
0 = 0 [This means all the value of x in the range (x - 3) ≤ 0 satisfy the equation ]
Hence, the rephrased question should be is x ≤ 3?

While solving absolute value problems do not forget the definition of absolute value.

Hope that helps.
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