SOLUTION: Is there a value of n such that n^2 + n + 1 is a multiple of 5?

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Question 711687: Is there a value of n such that n^2 + n + 1 is a multiple of 5?
Answer by Stitch(470) About Me  (Show Source):
You can put this solution on YOUR website!

Yes, there are values of n that produce multiples of 5. On the graph lines are drawn when Y=5, Y=10, Y=15 & Y=20. However the values of n that produce a multiple of 5 may not be whole numbers.
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For example we can solve for what value of n makes the equation equal 10.
n%5E2+%2B+n+%2B+1+=+10
Subtract 10 from both sides
n%5E2+%2B+n+-+9+=+0
Now we can use the quadratic equation to solve for n
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation an%5E2%2Bbn%2Bc=0 (in our case 1n%5E2%2B1n%2B-9+=+0) has the following solutons:

n%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A1%2A-9=37.

Discriminant d=37 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+37+%29%29%2F2%5Ca.

n%5B1%5D+=+%28-%281%29%2Bsqrt%28+37+%29%29%2F2%5C1+=+2.54138126514911
n%5B2%5D+=+%28-%281%29-sqrt%28+37+%29%29%2F2%5C1+=+-3.54138126514911

Quadratic expression 1n%5E2%2B1n%2B-9 can be factored:
1n%5E2%2B1n%2B-9+=+1%28n-2.54138126514911%29%2A%28n--3.54138126514911%29
Again, the answer is: 2.54138126514911, -3.54138126514911. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-9+%29

The quadratic shows us that when n is equal to -3.54 or 2.54, the equation is equal to 10, which is a multiple of five.