Question

The sum of the squares of three consecutive integers is 194. What are the integers ?

Answer 1

1st consecutive integer : x
2nd : x+1
3rd : x+2

(x)^2 + (x+1)^2 + (x+2)^2 = 194
x^2 + x^2 + 2x + 1 + x^2 + 4x + 4 = 194
3x^2 + 6x + 5 = 194
3x^2 + 6x - 189 = 0
3(x^2 + 2x - 63) = 0
(x+9)(x-7) = 0
x = -9 or x = 7


when x=-9
-9^2 + (-9+1)^2 + (-9+2)^2
= 81 + 64 + 49
=162 (reject)

when x=7
7^2 + (7+1)^2 + (7+2)^2
= 49 + 64 +81
= 194

so the three consecutive integers are 7,8,9 .
There hope this helps