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Bunuel
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If #x is defined for all x > 2 as the square root of the number that 16 May 2022, 08:45
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C
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85% (01:06) correct 15% (01:36) wrong based on 75 sessions

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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) ?

A. 0
B. 1
C. 2
D. 3
E. 4
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C
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RiyadhArifin237
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If #x is defined for all x > 2 as the square root of the number that 16 May 2022, 13:06
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Is it really math from sequence and funtion topic??

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wadhwakaran
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If #x is defined for all x > 2 as the square root of the number that 16 May 2022, 21:41
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#x= sqrt(x+2)
#7= sqrt(9)=3
#-1= sqrt(-1+2)=1
3-1=2
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If #x is defined for all x > 2 as the square root of the number that 16 May 2022, 23:44
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wadhwakaran
#x= sqrt(x+2)
#7= sqrt(9)=3
#-1= sqrt(-1+2)=1
3-1=2

Correct is 4.. we need some official solution from GMAT Club..

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If #x is defined for all x > 2 as the square root of the number that 17 May 2022, 04:59
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IMO 2
#x is defined for all x > –2 as the square root of the number that is 2 more than x
This essentially means
\(#x=\sqrt{(x+2)}\)
Few examples would be :
\(#3 = \sqrt{(3+2)}\) and \(#4 =\sqrt{(6)}\)
now we need to find #7 and #1
\(#7=\sqrt{(7+2)} = \sqrt{ 9} = 3\\
#-1=\sqrt{(-1+2)} = \sqrt{1 }= 1\\
\)
so, #7-#1= 3-1 =2
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If #x is defined for all x > 2 as the square root of the number that 19 Aug 2022, 11:25
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Given that #x is defined as the square root of the number that is 2 more than x and we need to find the value of #7 - #(-1)

#7 = square root of the number that is 2 more than 7
=> #7 = \(\sqrt{(7 + 2)}\) = \(\sqrt{9}\) = 3
Similarly, #(-1) = \(\sqrt{(-1 + 2)}\) = \(\sqrt{1}\) = 1

=> #7 - #(-1) = 3 - 1 = 2

So, Answer will be C
Hope it helps!

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If #x is defined for all x > 2 as the square root of the number that 22 Oct 2024, 13:30
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