SOLUTION: The sum of the squares of two consecutive positive intergers is 164 find the intergers,and please include the heading to define the variables.

Question 20110: The sum of the squares of two consecutive positive intergers is 164 find the intergers,and please include the heading to define the variables.
Found 2 solutions by wuwei96815, olayinka003:
Answer by wuwei96815(245) (Show Source): You can put this solution on YOUR website!
n=the first positive integer
n+2=the second positive integer

n^2 + n^2+2 = 164
n^2 + n^2 = 164-2
2n^2 = 162
n^2 = 162/2 = 91
n = 9
n+2 = 11

I hope that you are satisfied with this solution.
Answer by olayinka003(5) (Show Source): You can put this solution on YOUR website!
let x be the first positive integer
let y be the second positive integer
x^2 + n^2 + 2 = 164
x^2 + n^2 = 164 - 2
2n^2 = 162
n^2 = 162/2
n^2 = 81
squaring the both sides we have:
n = 9(first positive integer)
n^2 + 2 (second positive integer).
you ll see that n=9 and sqrt of 9 =81
n^2+ 2=81 + 2 =83
164
hope you got it
RELATED QUESTIONS
the sum of the squares of two consecutive positive even intergers is 340. find the... (answered by JimboP1977)
The sum of the squares of two consecutive positive intergers is 85. What are the... (answered by Paul)
The differnece between the squares of two consecutive positive odd intergers is 24. Find... (answered by checkley71)
My question: The sum of the squares of two consecutive positive intergers is 85; Find... (answered by wgunther)
The Sum of the squares of two consecutive positive intergers is 85.Find the... (answered by Earlsdon)
the sum of the squares of two consecutive odd positive intergers is 202. find the... (answered by solver91311,Alan3354)
Find two consecutive positive intergers the sum of whose squares is... (answered by tommyt3rd)
The Product of two consecutive positive intergers is 1 more than their sum. Find the... (answered by checkley77,Electrified_Levi)
The product of two consecutive positive intergers is 55 more than their sum. Find the... (answered by checkley77)