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Difficulty:
55%
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Question Stats:
63% (02:04) correct
37%
(01:47)
wrong
based on 213
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In the equation \(n^2 – kn + 16 = 0\), n is an integer. Which of the following could be the value of k?
I. 8
II. 15
III. –17
A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III
I. 8
II. 15
III. –17
A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III
Updated on: 06 Nov 2023, 06:08
Originally posted by $!vakumar.m on 23 Mar 2023, 02:42.
Last edited by $!vakumar.m on 06 Nov 2023, 06:08, edited 1 time in total.
Last edited by $!vakumar.m on 06 Nov 2023, 06:08, edited 1 time in total.
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n^2–kn+16=0; given n is an integer
I. k = 8
n^2 - 8n + 16 =0
(n-4)^2 = 0
=> n = 4 which is an integer. I is Correct
II. k = 15
n^2 - 15n + 16 = 0
sqrt(15^2 -4*1*16) = sqrt161 is not an integer, hence, n is not an integer. II is incorrect.
III. k = –17
n^2 - (-17)n + 16 = 0
n^2 + 17n + 16 = 0
(n+1) * (n+16) = 0
n = -1, n = -16. So n is an integer. III is correct.
Answer D
I. k = 8
n^2 - 8n + 16 =0
(n-4)^2 = 0
=> n = 4 which is an integer. I is Correct
II. k = 15
n^2 - 15n + 16 = 0
sqrt(15^2 -4*1*16) = sqrt161 is not an integer, hence, n is not an integer. II is incorrect.
III. k = –17
n^2 - (-17)n + 16 = 0
n^2 + 17n + 16 = 0
(n+1) * (n+16) = 0
n = -1, n = -16. So n is an integer. III is correct.
Answer D
