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# If #x is defined for all x > –2 as the square root of the nu

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GMAT Club Legend
Joined: 07 Jun 2014
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If #x is defined for all x > –2 as the square root of the nu [#permalink]  30 Jul 2018, 09:58
Expert's post
00:00

Question Stats:

66% (00:32) correct 33% (00:00) wrong based on 3 sessions
If $$#x$$ is defined for all x > –2 as the square root of the number that is 2 more than x, what is the value of $$#7 - #(-1)$$?

[Reveal] Spoiler: OA
2

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Sandy
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Intern
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Re: If #x is defined for all x > –2 as the square root of the nu [#permalink]  11 Aug 2018, 12:40
This question isn't too involved, but it's still good form, at least for practice, to write out what #x means so you have practice for a more involved version of the question on test day.

$$#x = \sqrt{x+2}$$

Then $$#7 = \sqrt{7+2} = \sqrt{9} = 3$$

and

$$#(-1) = \sqrt{-1+2} = \sqrt{1} = 1$$

So $$#7-#(-1) = 3-1 = 2$$
GMAT Club Legend
Joined: 07 Jun 2014
Posts: 4750
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 93

Kudos [?]: 1665 [0], given: 396

Re: If #x is defined for all x > –2 as the square root of the nu [#permalink]  11 Aug 2018, 16:35
Expert's post
Explanation

This function defines a made-up symbol, rather than using traditional notation such as f(x).

First, translate the function:

$$#x = \sqrt{x+2}$$

The square root of any value greater than or equal to 0 is the non-negative square root of the value.

That is, the square root of 4 is just +2, not –2. Thus:
$$#7 =\sqrt{7+2} =\sqrt{9} = 3$$
$$#(-1) = \sqrt{-1+2}= \sqrt{1}= 1$$
Finally, $$#7 - #(-1) = 3 - 1 = 2$$.
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Sandy
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Re: If #x is defined for all x > –2 as the square root of the nu   [#permalink] 11 Aug 2018, 16:35
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