SOLUTION: The squares of three consecutive odd numbers total 8 more than nine times the sum of the three numbers. Find the numbers?

Question 274505: The squares of three consecutive odd numbers total 8 more than nine times the sum of the three numbers. Find the numbers?
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
Let x, x+2 & x+4 be the 3 odd numbers.
x^2+(x+2)^2+(x+4)^2=9(x+x+2+x+4)+8
x^2+x^2+4x+4+x^2+8x+16=9(3x+6)+8
3x^2+12x+20=27x+54+8
3x^2+12x-27x+20-54-8=0
3x^2-15x-42=0
3(x^2-5x-14)=0
3(x-7)(x+2)=0
x-7=0
x=7 ans.
x+2=0
x=-2 ans.
Proof:
7^2+9^2+11^2=9(7+9+11)+8
49+81+121=9*27+8
251=243+8
251=251
-2^2+0^2+2^2=9(-2+0+2)+8
4+0+4=9*0+8
8=8
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