Question 119712: Find three consecutive integers such that the sum of their squares is 77.
Found 2 solutions by ankor@dixie-net.com, stanbon:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! x^2 + (x+1)^2 + (x+2)^2 = 77
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FOIL
x^2 + (x^2 + 2x + 1) + (x^2 + 4x + 4) = 77
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Group like terms on the left:
x^2 + x^2 + x^2 + 2x + 4x + 1 + 4 - 77 = 0
:
3x^2 + 6x - 72 = 0
:
Simplify, divide by 3
x^2 + 2x - 24 = 0
:
Factors to:
(x+6)(x-4) = 0
x = +4; then: 4, 5, 6 are the 3 consecutive digits
and
x = -6; then -6, -5, -4 are the 3 consecutive digits
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! Find three consecutive integers such that the sum of their squares is 77.
1st: x
2nd: x+1
3rd: x+2
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EQUATION:
x^2 + (x+1)^2 + (x+2)^2 = 77
3x^2 +6x +5 = 77
3x^2+6x-72=0
x^2+2x-24=0
(x+6)(x-4)=0
x = -6 or x=4
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If x = -6, then x+1 = -5, and x+2 = -4
If x = 4, then x+1 = 5, and x+2 = 6
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Cheers,
Stan H.
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