SOLUTION: Find 3 consecutive intergers such that the sum of their squares is 77.

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Question 148004: Find 3 consecutive intergers such that the sum of their squares is 77.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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Find 3 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 + x^2 + 2x + 4x + 1 + 4 = 77
:
3x^2 + 6x + 5 = 77
:
3x^2 + 6x + 5 - 77 = 0
:
3x^2 + 6x - 72 = 0
Simplify /3
x^2 + 2x - 24 = 0
Factor
(x+6)(x-4) = 0
Two solutions:
x=-6; you can consider this as a solution (they did not specify positive)
-6^2 + -5^2 + -4^2 = + 77
and
x=+4, (16 + 25 + 36 = 77)