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Pls explain this Algebra PS

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by theCodeToGMAT » Sat Oct 26, 2013 9:21 am
Solution 1: sqrt(x^2 -6x +9) + sqrt(2-x) + x - 3

sqrt((x-3)^2) + sqrt(2-x) + x - 3
x - 3 + sqrt(2-x) + x - 3
2x - 6 + sqrt(2-x)


SOlution 2: sqrt((3-x)^2) + sqrt(2-x) + x - 3
3 - x + sqrt(2-x) + x - 3
sqrt(2-x)

Since, question says.. that resultant under square root is greater then 0 .. So, SOLUTION 2 IS VALID
[spoiler]{A}[/spoiler]
What is the OA?
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by GMATGuruNY » Sat Oct 26, 2013 10:43 am
Let x=0.
Then √(x² - 6x + 9) + √(2-x) + x - 3 = √9 + √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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by [email protected] » Sat Oct 26, 2013 6:57 pm
Hey GmatGuru,

How did you start with assuming X as 0, why not another value?


Thanks


GMATGuruNY wrote:Let x=0.
Then √(x² - 6x + 9) + √(2-x) + x - 3 = √9 + √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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by theCodeToGMAT » Mon Oct 28, 2013 9:38 am
The possible values for "x" as per the question are: 0, 1 and 2 because question says that expression under the square root is either greater than or equal to 0.

So,

If X=1

(3-1) + sqrt(1) + 1 -3 = 2 + 1 + 1 - 3 = 1

{A} = sqrt(2-1) = 1

Also, if X=2

(3-2) + sqrt(0) + 2 - 3 = 0

{A} = sqrt(2-2) = 0
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by Matt@VeritasPrep » Mon Oct 28, 2013 10:21 am
This could be made a little bit clearer.

For any value of b, √(b²) = |b|

So

√(x²-6x+9) = √((x-3)²) = |x-3|

So

√(x²-6x+9) + √(2-x) + (x-3)
= |x-3| + √(2-x) + (x - 3)

Since (2-x) is nonnegative, (2-x) ≥ 0, or 2 ≥ x.

2 ≥ x implies that 0 > (x-3), which implies that |x-3| = -(x-3)

So |x-3| + √(2-x) + (x-3)
= -(x-3) + √(2-x) + (x-3)
= √(2-x)

And we're done!
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