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Tricky roots problem

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by tonebeeze » Wed Apr 20, 2011 1:48 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)

OA = A
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by vineeshp » Wed Apr 20, 2011 5:16 pm
I am not sure there is enough info in the question.

√(x^2 - 6x + 9) + √(2 - x) + x - 3.
√(x - 3)^2 + √(2 - x) + x - 3.

Now (x - 3)^2 >= 0 . This is already proven as it is a square.
(2 - x) >= 0 --> x <=2.

Now if the root of
√(x^2 - 6x + 9) is +- (x-3) > Both values are valid, as the square of this still satisfies the condition.

If root is - (x-3) . Then eqn comes down to - (x-3) + √(2 - x) + x - 3 = √(2 - x) . Answer will be A.

If root is (x-3), Then eqn comes down to - (x-3) + √(2 - x) + x - 3 = 2x - 6 + √(2 - x) . Answer will be B.

So what am I missin?
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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by bajwa2307 » Wed Apr 20, 2011 5:17 pm
you sure all the signs in the equation are correct? I think there should be a minus in there instead of a plus

x^2 - 6x + 9 can be written as (x -3)^2

so, √(x^2 - 6x + 9) can be written as √(x -3)^2 = x - 3


now is where the question might be wrong, if there is minus infront of the other x-3 such that it is written as -(x+3), the terms cancel out and we are left with the desired value √(2 - x)

can someone verify this please.
Verbal is testing my patience
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by MAAJ » Wed Apr 20, 2011 5:48 pm
I get (A) and (B) :/.... there most be something wrong with this problem. What's the source?

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

Positive abs
x - 3 + sqrt(2 - x) + x - 3
2x - 6 + sqrt(2 - x)

Negative abs
-x + 3 + sqrt(2 - x) + x - 3
sqrt(2 - x)
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by Anurag@Gurome » Thu Apr 21, 2011 9:38 pm
tonebeeze wrote: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)

OA = A
Solution:
Now all expressions under square root sign are positive or zero.
So, 2-x >= 0, or x <= 2.
Now x^2 - 6x + 9 = (x-3)^2.
So, sqrt(x^2 -6x+9) = (x-3) or -(x-3) = 3-x.
Now, in GMAT we just consider positive square root value.
Now, for x <= 2, x-3 < 0. So, just take 3-x as the square root.
Hence, the expression becomes 3-x +sqrt(2-x) + x-3 = sqrt(2-x).

The correct answer is a.
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by GMATGuruNY » Fri Apr 22, 2011 3:50 am
tonebeeze wrote: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)

OA = A
We can plug in a value for x.

Let x=1:
√(x² - 6x + 9) + √(2 - x) + x - 3
= √(1² - 6*1 + 9) + √(2-1) + 1 - 3
= √4 - 1
= 1. This is our target.

Now we plug x=1 into all the answers to which yields our target of 1.

Only answer choice A works:
√(2-x) = √(2-1) = 1.

The correct answer is A.
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