SOLUTION: Find two consecutive integers, such that the sum of their squares is 321

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Question 231946: Find two consecutive integers, such that the sum of their squares is 321
Answer by Alan3354(69443) About Me  (Show Source):
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Find two consecutive integers, such that the sum of their squares is 321
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x^2 + (x+1)^2 = 321
2x^2 + 2x + 1 = 321
2x^2 + 2x - 320 = 0
x^2 + x - 160 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B1x%2B-160+=+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=%281%29%5E2-4%2A1%2A-160=641.

Discriminant d=641 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+641+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%281%29%2Bsqrt%28+641+%29%29%2F2%5C1+=+12.1589889011722
x%5B2%5D+=+%28-%281%29-sqrt%28+641+%29%29%2F2%5C1+=+-13.1589889011722

Quadratic expression 1x%5E2%2B1x%2B-160 can be factored:
1x%5E2%2B1x%2B-160+=+%28x-12.1589889011722%29%2A%28x--13.1589889011722%29
Again, the answer is: 12.1589889011722, -13.1589889011722. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-160+%29

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No integer solution.