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In the equation n^2 – kn + 16 = 0, n is an integer.
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14 Mar 2019, 10:50
14 Mar 2019, 10:50
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42% (01:18) wrong based on 104 sessions
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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 \(k\)?
Indicate all such values
A. 8
B. 15
C. –17
Indicate all such values
A. 8
B. 15
C. –17
Re: In the equation n^2 – kn + 16 = 0, n is an integer.
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20 Mar 2019, 16:19
20 Mar 2019, 16:19
1
Expert Reply
This is a great one for picking numbers.
With just a bit of fiddling, it's not too difficult to make 8 work. When n = 4, then the equation works.
Now here's the thing with the bigger numbers. When k and n are equal, then the left is equal to 16: too large. If n is larger than k, then the left will also be too large. But if n is too much less than k, then kn will become more than 16 larger than n². So we really only have to check one or two values.
For example, when n = 14 and k = 15, then we get: 14² - (14*15) + 16 = 2, not quite.
But when we go down further, to n = 13 and k = 15, we get: 13² - (13*15) + 16 = -10.
And then as n decreases, the answer will just get lower and lower. So the equation never works.
We'll follow the same process with -17. If n = -16, then: (-16)² - (-16*-17) + 16 = 0. Boom.
With just a bit of fiddling, it's not too difficult to make 8 work. When n = 4, then the equation works.
Now here's the thing with the bigger numbers. When k and n are equal, then the left is equal to 16: too large. If n is larger than k, then the left will also be too large. But if n is too much less than k, then kn will become more than 16 larger than n². So we really only have to check one or two values.
For example, when n = 14 and k = 15, then we get: 14² - (14*15) + 16 = 2, not quite.
But when we go down further, to n = 13 and k = 15, we get: 13² - (13*15) + 16 = -10.
And then as n decreases, the answer will just get lower and lower. So the equation never works.
We'll follow the same process with -17. If n = -16, then: (-16)² - (-16*-17) + 16 = 0. Boom.
Re: In the equation n^2 – kn + 16 = 0, n is an integer.
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01 May 2020, 03:15
01 May 2020, 03:15
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For those who remember the quadratic formula, we can solve this easily.
Let quadratic equation be of the form ax + bx^2 + c = 0.
So a = 1, b = -k and c = 16.
The determinant term D = (b^2 - 4ac) has to be greater than 0, AND the sum of (-b +/- root(D)) should be even for the solution of n to be an integer.
Case 1: 8
D = 64 - 4(1)(16) = 0 (Perfect square)
So D +/- k = 0 +- 2 which is an even integer.
Case 2: 15
D = 225 - 64 = 161 (Not perfect square)
We don't need to go further since there is no way that solution of n would be an integer, with a rational number in the formula.
Case 3: -17
D = 289 - 64 = 225 (Perfect square)
So D +- k = 225 +- (-17) = 225 + 17 and 225 - 17 = 242 and 208 (Both even)
So this works.
Hence A, C.
For those not aware of the quadratic formula, use the number adjustment method pointed out in the above comment.
Let quadratic equation be of the form ax + bx^2 + c = 0.
So a = 1, b = -k and c = 16.
The determinant term D = (b^2 - 4ac) has to be greater than 0, AND the sum of (-b +/- root(D)) should be even for the solution of n to be an integer.
Case 1: 8
D = 64 - 4(1)(16) = 0 (Perfect square)
So D +/- k = 0 +- 2 which is an even integer.
Case 2: 15
D = 225 - 64 = 161 (Not perfect square)
We don't need to go further since there is no way that solution of n would be an integer, with a rational number in the formula.
Case 3: -17
D = 289 - 64 = 225 (Perfect square)
So D +- k = 225 +- (-17) = 225 + 17 and 225 - 17 = 242 and 208 (Both even)
So this works.
Hence A, C.
For those not aware of the quadratic formula, use the number adjustment method pointed out in the above comment.
