If the sum of the squares of three consecutives integers is 194, find the integers.?

Answer 1

Let n be the middle integer. Then, the other 2 are n -1 and n +1. It follows that

(n -1)^2 + n^2 + (n +1)^2 = n^2 - 2n + 1 + n^2 + n^2 + 2n + 1 = 3n^2 + 2 =194 => 3n^2 = 192 => n^2 = 64.

So, we have 2 solutions:

n = 8, which implies the 3 integers are 7 , 8 and 9, or

n = -8, which implies the 3 integers are -9, -8 and -7

Answer 2

let:
x = fist number
(x+1) = second number
(x+2) = third number

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 + 5 - 194 = 0
3x^2 + 6x - 189 = 0 --> Quadratic Equation

A Quadratic Equation is of the form: Ax^2 + Bx + C = 0
Coefficients: A = 3, B = 6, C = -189


Determine x using the Quadratic Formula (see link below):

x = 7 --> 1st number
(x+1) = (7+1) = 8 ---> 2nd number
(x+2) = (7+2) = 9 ---> 3rd number

Checking:
(7)^2 + (8)^2 + (9)^2 = 49 + 64 + 81 = 194

Answer 3

7, 8 and 9

The way I did it was to divide the result by 3, find the square root of that number then take the nearest integer as the middle one of the three.

(194/3 = 64.6666)
(sq rt 64.6666 = 8.04)

Probably not the correct mathematical way to solve it, but it worked.