SOLUTION: The sum of the squares of two consecutive integers is 613. What are the integers? Let x be one integer. Therefore x +1 is the next integer. x^2+(x+1)^2=613

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The sum of the squares of two consecutive integers is 613. What are the integers? Let x be one integer. Therefore x +1 is the next integer. x^2+(x+1)^2=613      Log On


   

Question 324085: The sum of the squares of two consecutive integers is 613. What are the integers?
Let x be one integer. Therefore x +1 is the next integer.
x^2+(x+1)^2=613
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2B%28x%2B1%29%5E2=613
Complete the square.
x%5E2%2B%28x%5E2%2B2x%2B1%29=613
2x%5E2%2B2x%2B1=613
2%28x%5E2%2Bx%29=612
2%28x%5E2%2Bx%2B1%2F4%29=612%2B1%2F2
2%28x%2B1%2F2%29%5E2=1224%2F2%2B1%2F2
%28x%2B1%2F2%29%5E2=1225%2F4
x%2B1%2F2=0+%2B-+35%2F2
x=-1%2F2+%2B-+35%2F2
x=-36%2F2=-18 and x=34%2F2=17
.
.
.
The integers are either -18 and -17 or 17 and 18.