Question

the sum of square of two consecutive positive integer is 221. find the integer

Answer 1

the no. are

10 and 11
100+121
221

Answer 2

Let the consecutive integers be x and x + 1.

Given that the sum of two consecutive positive integers = 221.

x^2 + (x + 1)^2 = 221

x^2 + x^2 + 1 + 2x = 221

2x^2 + 2x + 1 - 221 = 0

2x^2 + 2x - 220 = 0

x^2 + x - 110 = 0

x^2 - 10x + 11x - 110 = 0

x(x - 10) + 11(x - 10) = 0

(x +11)(x - 10) = 0

x = 10 or -11.

x cannot be -ve, so x = 10.


Verification:

x^2 + (x + 1)^2 = 221

10^2 + 11^2 = 221

100 + 121 = 221

221 = 221.


Hope this helps!