For how many natural numbers n is the fraction 5n^2-9/2n+6 also a natural number?
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Answers
Given : (5n² - 9)/(2n + 6) is a natural number
n is also natural number
To Find : How many natural numbers n satisfy
Solution:
(5n² - 9)/(2n + 6) = k
=> (5n² - 9)/(n + 3) = 2k
Hence (5n² - 9)/(n + 3) is an even number
5n - 15
n + 3 _| 5n² - 9 |_
5n² + 15n
________
-15n - 9
-15n - 45
_______
36
(5n² - 9)/(n + 3) = 5n - 15 + 36/(n + 3)
36/(n + 3) is a natural number
n is a natural number
Hence n = 1 , 3 , 6 , 9 , 15 , 33
5n - 15 + 36/(n + 3)
n = 1 => 5n - 15 + 36/(n + 3) = - 1 hence not possible
n = 3 => 5n - 15 + 36/(n + 3) = 6 also even hence possible
n = 6 => 5n - 15 + 36/(n + 3) = 19 not even number hence not possible
n = 9 => 5n - 15 + 36/(n + 3) = 33 not even number hence not possible
n = 15 => 5n - 15 + 36/(n + 3) = 62 also even hence possible
n = 33 => 5n - 15 + 36/(n + 3) = 151 not even number hence not possible
Hence n = 3 and n = 15
2 possible values of n
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Answer: 3
Given : (5n² - 9)/(2n + 6) is a natural number
n is also natural number
To Find : How many natural numbers n satisfy
Solution:
(5n² - 9)/(2n + 6) = k
=> (5n² - 9)/(n + 3) = 2k
Hence (5n² - 9)/(n + 3) is an even n 5n - 15
n + 3 _| 5n² - 9 |_
5n² + 15n
________
-15n - 9
-15n - 45
_______
36
(5n² - 9)/(n + 3) = 5n - 15 + 36/(n + 3)
36/(n + 3) is a natural number
n is a natural number
Hence n = 1 , 3 , 6 , 9 , 15 , 33
5n - 15 + 36/(n + 3)
n = 1 => 5n - 15 + 36/(n + 3) = - 1 hence not possible
n = 3 => 5n - 15 + 36/(n + 3) = 6 also even hence possible
n = 6 => 5n - 15 + 36/(n + 3) = 19 not even number hence not possible
n = 9 => 5n - 15 + 36/(n + 3) = 33 not even number hence not possible
n = 15 => 5n - 15 + 36/(n + 3) = 62 also even hence possible
n = 33 => 5n - 15 + 36/(n + 3) = 151 not even number hence not possible
Hence n = 3 and n = 15
2 possible values of n
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