SOLUTION: Consecutive integers. I need to find three consecutive integers such that the sum of their squares is 77.

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Question 39885: Consecutive integers.
I need to find three consecutive integers such that the sum of their squares is 77.
Answer by vidhyak(98) About Me  (Show Source):
You can put this solution on YOUR website!
Let the 3 integers be x, x+1, x+2
x^2 + (x+1)^2 + (x+2)^2 = 77
x^2 + x^2 + 2x + 1 + x^2 + 4x + 4 = 77
3x^2 + 6x + 5 = 77
3x^2 + 6x - 72 = 0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 3x%5E2%2B6x%2B-72+=+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=%286%29%5E2-4%2A3%2A-72=900.

Discriminant d=900 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-6%2B-sqrt%28+900+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%286%29%2Bsqrt%28+900+%29%29%2F2%5C3+=+4
x%5B2%5D+=+%28-%286%29-sqrt%28+900+%29%29%2F2%5C3+=+-6

Quadratic expression 3x%5E2%2B6x%2B-72 can be factored:
3x%5E2%2B6x%2B-72+=+3%28x-4%29%2A%28x--6%29
Again, the answer is: 4, -6. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B6%2Ax%2B-72+%29



Solving you get 4, -6
The 3 integers are 4,5,6 or -6,-5,-4