For how many integers n is the fraction (2n^2+1)/(n^2-1)
also an integer?
Answers
Given : fraction (2n²+1)/(n²-1)
To Find : For how many integers n , given fraction is integer
Solution:
(2n²+1)/(n²-1)
= (2n²+1-2+2)/(n²-1)
= (2n² - 2 + 3)/(n² - 1)
= (2(n² -1) + 3)/(n² - 1)
= 2 + 3/(n² - 1)
3/(n² - 1) is integer if
n² - 1 = -1 , 1 , -3 or 3
Checking
n² - 1 = - 1 =>n² = 0 => n = 0
n² - 1 = 1 => n =± √2 ( not an integer)
n² - 1 = -3 => n² = -2 hence n is not real
n² - 1 = 3 => n² = 4 => n =± 2
So possible integral values of n are -2 , 0 , 2
Hence 3 values of n satisfy this
Verification :
(2n²+1)/(n²-1)
n = 0 => 1/-1 = - 1
n = -2 => 9/3 = 3
n = 2 => 9/3 = 3
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Answer:
Step-by-step explanation:
Given the fraction (2n^2 + 1)/(n^2 - 1), we need to find the values of n for which the numerator (2n^2 + 1) is divisible by the denominator (n^2 - 1) without any remainder.
To simplify the fraction, we can factorize the numerator and denominator:
Numerator: 2n^2 + 1 = (n^2 - 1) + (n^2 + 2)
Denominator: n^2 - 1 = (n + 1)(n - 1)
Now, we can cancel out the common factor (n^2 - 1) from the numerator and denominator:
(2n^2 + 1)/(n^2 - 1) = (n^2 + 2)/((n + 1)(n - 1))
For the fraction to be an integer, the numerator (n^2 + 2) must be divisible by both (n + 1) and (n - 1).
Let's check the possible integer values of n:
When n = -1:
Numerator = (-1)^2 + 2 = 3
Denominator = (-1 + 1)(-1 - 1) = 0
The fraction is undefined.
When n = 0:
Numerator = (0)^2 + 2 = 2
Denominator = (0 + 1)(0 - 1) = -1
The fraction is undefined.
When n = 1:
Numerator = (1)^2 + 2 = 3
Denominator = (1 + 1)(1 - 1) = 0
The fraction is undefined.
When n = 2:
Numerator = (2)^2 + 2 = 6
Denominator = (2 + 1)(2 - 1) = 3
The fraction is not an integer.
Therefore, there are no integer values of n for which the fraction (2n^2 + 1)/(n^2 - 1) is an integer.