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

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)   (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(75887)   (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(2933)   (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+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

77=77 checks

this checks too.
Happy Calculating!!!
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