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# M08-16

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Intern
Joined: 04 May 2017
Posts: 1

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01 Nov 2017, 07:23
I think this is a high-quality question and I agree with explanation.
Manager
Joined: 01 Feb 2017
Posts: 79

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09 Nov 2017, 00:51
Using the plug-in method is a faster way to get correct answer.

As we know that expression under square root is <=0, x must be <=2.

Plug in any value of x that is <=2, say x=1, in given expression​and each answer choice.

Only Choice A will match.

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Intern
Joined: 24 Jun 2013
Posts: 1

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23 Nov 2017, 03:28
Bunuel wrote:
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. $$x + \sqrt{x - 2}$$

Hie Bunuel,

Math Expert
Joined: 02 Sep 2009
Posts: 43896

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23 Nov 2017, 04:28
prk0789 wrote:
Bunuel wrote:
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. $$x + \sqrt{x - 2}$$

Hie Bunuel,

8. Exponents and Roots of Numbers

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Intern
Status: Working my way to the MBA
Joined: 19 Nov 2017
Posts: 2
Location: Brazil
WE: Consulting (Consulting)

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11 Feb 2018, 13:14
1
KUDOS
I think this is a high-quality question and I agree with explanation. There is a grammar error on the explanation. Instead of "than" there should be "then" before ":"
Math Expert
Joined: 02 Sep 2009
Posts: 43896

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11 Feb 2018, 20:49
rodmm wrote:
I think this is a high-quality question and I agree with explanation. There is a grammar error on the explanation. Instead of "than" there should be "then" before ":"

Edited the typo. Thank you.
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Intern
Joined: 16 Feb 2018
Posts: 1

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20 Feb 2018, 01:00
I think this is a high-quality question and I agree with explanation.
Re M08-16   [#permalink] 20 Feb 2018, 01:00

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# M08-16

Moderators: chetan2u, Bunuel

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