SOLUTION: Find two consecutive integers such that the sum of their squares is 61

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Question 589191: Find two consecutive integers such that the sum of their squares is 61
Found 3 solutions by AnlytcPhil, mathhelp@, Alan3354:
Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!
Find two consecutive integers such that the sum of their squares is 61
Smaller consecutive integer = N
Larger consecutive integer N+!

sum of their squares is 61

          N² + (N+1)² = 61

      N² + (N+1)(N+1) = 61

N² + (N² + N + N + 1) = 61

     N² + N² + 2N + 1 = 61

         2N² + 2N + 1 = 61

Get 0 on the right:

        2N² + 2N - 60 = 0

Divide every term by 2

         N² +  N - 30 = 0
 
Factor:

       (N + 6)(N - 5) = 0

Use zero factor principle:

      N + 6 = 0;    N - 5 = 0
          N = -6        N = 5



Using answer N = -6
Smaller consecutive integer = N = -6
Larger consecutive integer N+1 = -6+1 = -5

Using answer N = 5
Smaller consecutive integer = N = 5
Larger consecutive integer N+1 = 5+1 = 6

Two answers:  -6 and -5,   or 5 and 6

Edwin

Answer by mathhelp@(10)   (Show Source): You can put this solution on YOUR website!
Simple
use x as a variable
you are going to solve for x, so do this:
x^2+(x+1)^2=61
so its x^2+x^2+2x+1=61
2x^2+ 2x + 1 =61
quadratic formula

b=2
a=2
c=1
plug it in and you are done!





















If you get these answers, then you are correct:
5 and 6

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 61
----------
61/2 = 30.5
apx, the center
--> 5 & 6
Also -6 & -5

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