SOLUTION: how do i find three consecutive intergers such that the sum of the squares is 77.

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Question 57584: how do i find three consecutive intergers such that the sum of the squares is 77.
Found 3 solutions by checkley71, stanbon, funmath:
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+(x+1)^2+(x+2)^2=77
x^2+x^2+2x+1+x^2+4x+4=77
3x^2+6x+5-77=0
3x^2+6x-72=0
x^2+2x-24=0
(x+6)(x-4)=0
x+6=0
x=-6
x-4=0
x=4 thus the other two numbers are 4+1=5 & 4+2=6
proof
4^2+5^2+6^2=77
16+25+36=77
77=77

Answer by stanbon(48558) About Me  (Show Source):
You can put this solution on YOUR website!
how do i find three consecutive intergers such that the sum of the squares is 77.
Let the consecutive integers be x, x+1, x+2.
EQUATION:
x^2+(x^1)^2+(x+2)^2=77
x^2+x^2+2x+1 + x^2+4x+2=77
3x^2+6x+3=77
3x^2+6x-74=0
x=[-6+-sqrt(6^2-4*3*-74)]/6
x=[-6+-sqrt(924)]/6
Comment: This answer is not an integer.
There are no consecutive integers that meet
the requirement of the equation.
Cheers,
Stan H.

Answer by funmath(2873) About Me  (Show Source):
You can put this solution on YOUR website!
how do i find three consecutive intergers such that the sum of the squares is 77.
Let the 1st integer=x
then the 2nd integer=x+1
then the third integer=x+1+1=x+2
Then the equation to solve is:
x%5E2%2B%28x%2B1%29%5E2%2B%28x%2B2%29%5E2=77
x%5E2%2B%28x%5E2%2B2x%2B1%29%2B%28x%5E2%2B4x%2B4%29=77
x%5E2%2Bx%5E2%2Bx%5E2%2B2x%2B4x%2B1%2B4=77
3x%5E2%2B6x%2B5=77
3x%5E2%2B6x%2B5-77=0
3x%5E2%2B6x-72=0
3%28x%5E2%2B2x-24%29=0
%283%2F3%29%28x%5E2%2B2x-24%29=0%2F3
%28x%5E2%2B2x-24%29=0
%28x%2B6%29%28x-4%29=0
x+6=0 and x-4=0
x+6-6=0-6 and x-4+4=0+4
x=-6 and x=4
There are two possible sets of integers: -6,-5,-4 and 4,5,6
:
Check by seeing if the sum of their squares is = to 77
%28-6%29%5E2%2B%28-5%29%5E2%2B%28-4%29%5E2=77
36%2B25%2B16=77
77=77 checks
%284%29%5E2%2B%285%29%5E2%2B%286%29%5E2=77
16%2B25%2B36=77
77=77 this checks too.
Happy Calculating!!!