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)   (Show Source): You can put this solution on YOUR website!
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 (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=641 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 12.1589889011722, -13.1589889011722. Here's your graph:

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No integer solution.
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