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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Absolute values and inequalities
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topspin360
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topspin360
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theCodeToGMAT
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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]
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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GMATGuruNY
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Let x=0.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)
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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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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