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06 Nov 2023, 09:29
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C
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Difficulty:
95%
(hard)
Question Stats:
31% (02:15) correct
69%
(02:24)
wrong
based on 145
sessions
History
Date
Time
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Not Attempted Yet
For how many integer values of n is n^2 + 6n - 8 the square of an integer?
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
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23 Jan 2024, 02:25
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For how many integer values of \(n\) is \(n^2 + 6n - 8\) the square of an integer?
A. 0
B. 1
C. 2
D. 3
E. 4
We need to find how many integer values of \(n\) exist such that \(n^2 + 6n - 8=m^2\), for some integer \(m\).
Complete the square for the expression on the left-hand side:
Re-arrange:
Apply the difference of squares formula:
So, we find that the product of two integers, \((n + 3 - m)\) and \((n + 3 + m)\), equals the prime number 17. 17 can be expressed as the product of two integers in only two ways: \(17 = 1*17\) and \(17 = (-1)(-17)\).
Therefore, there are two such values of \(n\) for which \(n^2 + 6n - 8\) is the square of an integer.
Answer: C
A. 0
B. 1
C. 2
D. 3
E. 4
We need to find how many integer values of \(n\) exist such that \(n^2 + 6n - 8=m^2\), for some integer \(m\).
Complete the square for the expression on the left-hand side:
- \(n^2 + 6n + 9 - 17=m^2\)
\((n + 3)^2 - 17=m^2\)
Re-arrange:
- \((n + 3)^2 - m^2 = 17\)
Apply the difference of squares formula:
- \((n + 3 - m)(n + 3 + m) = 17\)
So, we find that the product of two integers, \((n + 3 - m)\) and \((n + 3 + m)\), equals the prime number 17. 17 can be expressed as the product of two integers in only two ways: \(17 = 1*17\) and \(17 = (-1)(-17)\).
- \((n + 3 - m)=1\) and \((n + 3 + m) =17\) or vice versa gives \(n=6\).
\((n + 3 - m)=-1\) and \((n + 3 + m) =-17\) or vice versa gives \(n=-12\).
Therefore, there are two such values of \(n\) for which \(n^2 + 6n - 8\) is the square of an integer.
Answer: C
General Discussion
06 Nov 2023, 13:24
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Bunuel
Let \(k =\) integer
\(n^2 + 6n - 8 = k^2\)
\(n^2 + 6n + 9 - 17 = k^2\)
\((n+3)^2 - 17 = k^2\)
\((n+3)^2 - k^2 = 17\)
\((n+3 - k)(n+3+k) = 17\)
As 17 is a prime number, the two parts of the product must be \(1\) & \(17\), \(-1\) & \(-17\)
Case 1
\(n + 3 - k = 1\)
\(n + 3 + k = 17\)
Adding both equations we get -
\(2n + 6 = 18\)
\(2n = 12\)
\(n = 6\)
Case 2
\(n + 3 - k = -1\)
\(n + 3 + k = -17\)
Adding both equations we get -
\(2n + 6 = -18\)
\(2n = -24\)
\(n = -12\)
Hence, for \(2\) values of \(n\), \(n^2 + 6n - 8\) is the square of an integer
Option C
P.S. - Great Question
