Algebra (hard one in my opinion)

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Algebra (hard one in my opinion)

by mariofelixpasku » Fri Oct 19, 2012 12:59 pm
Is the square root of (x-5)2 = 5-x?
(1) -x|x| > 0
(2) 5 - x > 0


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by GMATGuruNY » Fri Oct 19, 2012 5:23 pm
mariofelixpasku wrote:Is √(x-5)² = 5-x?
(1) -x|x| > 0
(2) 5 - x > 0
It helps to know the following:
Thus, √(x²) = |x|.
|x-y| = the DISTANCE between x and y.

In the problem at hand:
√(x-5)² = |x-5| = the DISTANCE between x and 5.
A distance must be greater than or equal to 0.

5-x = the DIFFERENCE between 5 and x.
A difference can be negative, 0, or positive.

The DIFFERENCE between two values will be equal to the DISTANCE between the two values whenever the DIFFERENCE is greater than or equal to 0.

Thus, |x-5| = 5-x whenever 5-x≥0.
Question rephrased: Is x≤5?

Statement 1: -x|x| > 0
Since |x| cannot be negative, both factors (-x and |x|) must be positive.
-x > 0
Since x<0, we know that x≤5.

Statement 2: 5-x > 0
Thus, x<5.

The correct answer is D.
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by prakhar bumb » Sat Oct 20, 2012 12:49 pm
I dont understand, how can -x|x| > 0 as how a negative no. multiplied by a positive no. be greater than 0

Also, could you please tell about the source of this problem