SOLUTION: Twice the square of an integer is 21 more than eleven times the integer. Find the integer.

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Question 144460: Twice the square of an integer is 21 more than eleven times the integer. Find the integer.
Answer by Alan3354(21610) About Me  (Show Source):
You can put this solution on YOUR website!
2I%5E2+=+11I+%2B+21
2I%5E2+-11I+-21+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-11x%2B-21+=+0) has the following solutons:

x%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=%28-11%29%5E2-4%2A2%2A-21=289.

Discriminant d=289 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--11%2B-sqrt%28+289+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-11%29%2Bsqrt%28+289+%29%29%2F2%5C2+=+7
x%5B2%5D+=+%28-%28-11%29-sqrt%28+289+%29%29%2F2%5C2+=+-1.5

Quadratic expression 2x%5E2%2B-11x%2B-21 can be factored:
2x%5E2%2B-11x%2B-21+=+%28x-7%29%2A%28x--1.5%29
Again, the answer is: 7, -1.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-11%2Ax%2B-21+%29


-1.5 is not an integer, so the answer is 7.
2 times 7 times 7 = 11 times 7 + 21