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GRE Prep Club Legend  Joined: 07 Jun 2014
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GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
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If #x is defined for all x > –2 as the square root of the nu [#permalink]
Expert's post 00:00

Question Stats: 75% (00:36) correct 25% (00:00) wrong based on 4 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
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Sandy
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Intern Joined: 10 Aug 2018
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Re: If #x is defined for all x > –2 as the square root of the nu [#permalink]
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 $$#7-#(-1) = 3-1 = 2$$
GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4857
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 105

Kudos [?]: 1783 , given: 397

Re: If #x is defined for all x > –2 as the square root of the nu [#permalink]
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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Try our free Online GRE Test 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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