SOLUTION: consecutive integers . find three consecutive integers such that the sum of their square is 77

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Question 154672: consecutive integers . find three consecutive integers such that the sum of their square is 77
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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The 3 consecutive integers: x, x+1, x+2
:
find three consecutive integers such that the sum of their squares is 77
x^2 + (x+1)^2 + (x+2)^2 = 77
FOIL
x^2 + (x^2 + 2x + 1) + (x^2 + 4x + 4) = 77
:
x^2 + x^2 + 2x + 1 + x^2 + 4x + 4 = 77;
Group like terms
x^2 + x^2 + x^2 + 2x + 4x + 1 + 4 = 77
:
3x^2 + 6x + 5 = 77
:
3x^2 + 6x + 5 - 77 = 0
:
3x^2 + 6x - 72 = 0
Simplify divide by 3
x^2 + 2x - 24 = 0
Factor
(x+6)(x-4) = 0
Two solutions:
x = -6 and x = +4
:
Our consecutive numbers:
-6, -5, -4
and
4, 5, 6