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805+ (Hard)|   Arithmetic|      
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Ravixxx
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If x is a positive real number, how many possible integer values are t 07 May 2020, 01:00
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

41% (02:06) correct 59% (02:38) wrong based on 146 sessions

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If x is a positive real number, how many possible integer values are there for \(\sqrt{ x^{2}+14x}-x\) ?

(A) 2
(B) 4
(C) 6
(D) 8
(E) 10
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C
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GMATinsight
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If x is a positive real number, how many possible integer values are t 07 May 2020, 01:35
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Ravixxx
If x is a positive real number, how many possible integer values are there for \(\sqrt{ x^{2}+14x}-x\) ?

(A) 2
(B) 4
(C) 6
(D) 8
(E) 10

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\(\sqrt{ x^{2}+14x}-x\) ?

\(\sqrt{ x^{2}+14x} = x\) ?

Squaring both sides

\(x^2 + 14x = (x+ Integer)^2\)

Let, \(Integer = I\)

\(x^2 + 14x = x^2+ Integer^2 + 2*x*Integer\)

i.e. \(2x (7-I) = I^2\)

'\(I\)' can take any value from 1 to 6

Answer: Option C
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If x is a positive real number, how many possible integer values are t 06 Jan 2021, 11:44
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Ravixxx
If x is a positive real number, how many possible integer values are there for \(\sqrt{ x^{2}+14x}-x\) ?

(A) 2
(B) 4
(C) 6
(D) 8
(E) 10

Kudos if you appreciate this question

\(\sqrt{ x^{2}+14x}-x\) ?

\(\sqrt{ x^{2}+14x} = x\) ?

Squaring both sides

\(x^2 + 14x = (x+ Integer)^2\)

Let, \(Integer = I\)

\(x^2 + 14x = x^2+ Integer^2 + 2*x*Integer\)

i.e. \(2x (7-I) = I^2\)

'\(I\)' can take any value from 1 to 6

Answer: Option C
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Quick question - Why can't I = 0?
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If x is a positive real number, how many possible integer values are t 06 Jan 2021, 21:49
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aalam383
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Ravixxx
If x is a positive real number, how many possible integer values are there for \(\sqrt{ x^{2}+14x}-x\) ?

(A) 2
(B) 4
(C) 6
(D) 8
(E) 10

Kudos if you appreciate this question

\(\sqrt{ x^{2}+14x}-x\) ?

\(\sqrt{ x^{2}+14x} = x\) ?

Squaring both sides

\(x^2 + 14x = (x+ Integer)^2\)

Let, \(Integer = I\)

\(x^2 + 14x = x^2+ Integer^2 + 2*x*Integer\)

i.e. \(2x (7-I) = I^2\)

'\(I\)' can take any value from 1 to 6

Answer: Option C

Quick question - Why can't I = 0?
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aalam383 X would also have to be 0 for that. And X is given to be a "positive real number"
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If x is a positive real number, how many possible integer values are t 19 Jun 2021, 01:44
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If x is a positive real number, how many possible integer values are there for \(\sqrt{ x^{2}+14x}-x\) ?

\(\sqrt{ x^{2}+14x}-x\) = Integer value

Or, \(x^{2}+14x = (x + I)^{2}\)

or, 14x -2*Ix = Integer or, 2x (7- I) = Integer^2 = Positive Integer, as x >0 so, 7-I >0 and I = 1,2,3,4,5,6.

So, I think C. :)
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If x is a positive real number, how many possible integer values are t 17 Feb 2024, 18:08
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