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Absolute values and inequalities

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Absolute values and inequalities

by topspin360 » Sat Jan 04, 2014 9:01 am
Hello,

Can someone please solve the following problem in detail? Especially the part where we have to evaluate absolute value based on two cases: x < 0 and x > 0.

If each expression under the square root is greater than or equal to 0, what is Sqrt(x^2 - 6x + 9) + Sqrt(2 - x) + x - 3?

A) Sqrt (2 - x)
B) 2x - 6 + Sqrt(2 - x)
C) Sqrt(2 - x) + x - 3
D) 2x - 6 + Sqrt(x - 2)
E) x + Sqrt(x - 2)

Thanks!
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by topspin360 » Sat Jan 04, 2014 9:01 am
OA is A
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by theCodeToGMAT » Sat Jan 04, 2014 10:42 am
Sqrt(x^2 - 6x + 9) + Sqrt(2 - x) + x - 3

sqrt((x-3)(x-3)) + sqrt(2-x) + (x-3)

Since, we know that term under sqrt >= 0

2 - x >= 0

x <=2
this implies that sqrt((x-3)(x-3)) is rather sqrt((3-x)(3-x)) because (x-3) will result in negative term, although (x-3)(x-3) will result in positive resultant.

hence, equation turns to:
sqrt(3-x)^2 + sqrt(2-x) + (x-3)
= 3-x + x-3 + sqrt(2-x)
= sqrt(2-x)

[spoiler]{A}[/spoiler]
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by GMATGuruNY » Sat Jan 04, 2014 9:01 pm
If each expression under the square root is greater than or equal to 0, what is √(x^2 - 6x + 9) + √(2 - x) + x - 3?

a. √(2-x)
b. 2x - 6 + √(2-x)
c. √(2-x) + x - 3
d. 2x - 6 + √(x-2)
e. x + √(x-2)
Let x=0.
Then √(x^2 - 6x + 9) + √(2 - x) + x - 3 = 3 + √2 + 0 - 3 = √2. This is our target.
Now plug x=0 into the answers to see which yields our target of √2.
Only A works:
√(2-x) = √(2-0) = √2.

The correct answer is A.
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