Solve the problem
-The sum of the squares of two consecutive integers is 221. Find the integers.
A) $10 , - 11$
B) 10,11 or $- 10 , - 11$
C) 10,11
D) $- 10 , - 11$

Question

Solve the problem
-The sum of the squares of two consecutive integers is 221. Find the integers. 
A) $10 , - 11$
B) 10,11 or $- 10 , - 11$
C) 10,11
D) $- 10 , - 11$

Quizplus · Accepted Answer

The equation for the sum of the squares of two consecutive integers $n$ and $n+1$ is $n^2 + (n+1)^2 = 221$. Simplifying, we get $2n^2 + 2n - 220 = 0$, which simplifies further to $n^2 + n - 110 = 0$. Factoring, we find $n = 10$ or $n = -11$. Therefore, the pairs are (10, 11) and (-11, -10), which corresponds to choice B.

Solve the Problem -The Sum of the Squares of Two 10,−1110 , - 1110,−11

Solve the Problem
-The Sum of the Squares of Two $10 , - 11$