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sqrt [(x-3)^2] = 3-x ?

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Re: sqrt [(x-3)^2] = 3-x ?

by tritrantran » Thu Dec 11, 2008 11:30 am
vittalgmat wrote:Is sqrt [(x-3)^2] = 3-x

1) x is not equal to 3
2) -x*lxl > 0

not sure of the OA B

Thanks
From the question we know that 3 -x can't be negative since we can't take a sqrt of a negative number, so x has to be less than or equal to 3 for the equation to work.

1) x is not equal to 3

x = 4
doesn't work since we can't take the sqrt of a negative value

x = -4
will work

INSUFFICIENT

2) -x*lxl > 0

this tells us that -x*positive > 0

so - x must be positive, and x must be negative

if x is negative, then any value will make the original equation true.

SUFFICIENT B)
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Re: sqrt [(x-3)^2] = 3-x ?

by parallel_chase » Thu Dec 11, 2008 11:31 am
vittalgmat wrote:Is sqrt [(x-3)^2] = 3-x

1) x is not equal to 3
2) -x*lxl > 0

not sure of the OA B

Thanks
Rule: sqrtx^2 = lxl

sqrt [(x-3)^2] = 3-x

lx-3l = 3-x

to satisfy above equation x can be any negative value, 0,1,2 & 3.

Statement I
1) x is not equal to 3
x could 6,7,8 or any negative value. We can have both yes & no. Insufficient.

Statement II
2) -x*lxl > 0

for this equation to hold true x has to be negative, therefore any negative value will give us the answer yes. Sufficient.

Hence B.

Hope this helps.
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by vittalgmat » Thu Dec 11, 2008 2:51 pm
thanks everyone. got it.
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by ronniecoleman » Thu Dec 11, 2008 10:44 pm
IMO B
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by chufus » Mon Dec 05, 2011 4:53 am
Guys. I still don't get it. The absolute method |x-3| works but lets plug in a number and see. Statement B only tells us that x is negative for sure.

Let x = -7

So equation becomes

Sqrt (-7-1)^2 = 3 - (-7)

so sqrt (100) = 10

Now sqrt (100) can have two values, 10 and -10 .. The other side is clearly 10.

So how is B the correct answer ?
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by neelgandham » Mon Dec 05, 2011 5:25 am
Square root of a number x^2 is defined as below(You can Wiki or Google if you still doubt this)
f(x) = Square root(x^2) = x if x >= 0 and -x if x <0. In other words, f(x) = Square root(x^2) = |x|(always positive).

Square root(100) can be 10 and never -10.

sqrt [(x-3)^2] = 3-x = -(x-3), only if x-3 is negative. The question can now be rephrased to
Is (x-3)<0 ?
1) x is not equal to 3
The value of x can be greater than 3(x-3 >0) or less than 3 (x-3<0). Hence Insufficient!
2) -x*|x| > 0
-x*|x| > 0 Implies -x * Positive number = Positive number. So -x should be a positive number and therefore x a negative number.If x is a negative number, x-3 is also a negative number(-ve + -ve = -ve).So, we can answer the question, Is (x-3)<0 ? with a YES! Hence sufficient!

Answer B
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by GMATGuruNY » Mon Dec 05, 2011 9:03 am
vittalgmat wrote:Is sqrt [(x-3)^2] = 3-x

1) x is not equal to 3
2) -x*|x| > 0

not sure of the OA B
Be definition:
√(x²) = |x|.
|x-y| is the DISTANCE between x and y.
The DISTANCE between two numbers must be greater than or equal to 0.

The question stem above, rephrased: Is |x-3| = 3-x?
In words:
Is the DISTANCE between x and 3 equal to the DIFFERENCE of 3 and x?
The answer will be YES if the DIFFERENCE of 3 and x is greater than or equal to 0:
3-x≥0
x≤3.

The question stem rephrased: Is x≤3?

Statement 1: x is not equal to 3.
It is possible that x<3 or that x>3.
INSUFFICIENT.

Statement 2: -x*|x| > 0 .
Thus, the left-hand side must be positive*positive or negative*negative.
Since |x| cannot be negative, both factors on the left-hand side must be positive.
Thus:
-x>0
x<0.
Since x<0, we know that x≤3.
SUFFICIENT.

The correct answer is B.
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