SOLUTION: The sum of the squares of three consecutive even integers is 116. Find the integers... This is what I have tried... x + (x + 2)+ (x + 4)= 116 3x + 6 = 116 -6

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Question 158814: The sum of the squares of three consecutive even integers is 116. Find the integers...
This is what I have tried...
x + (x + 2)+ (x + 4)= 116
3x + 6 = 116
-6 - 6
3x = 110
3x/3 110/3
x = 36.6666
This doesnt make sense. What am i doing wrong?

Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website!
You are asked for the sum of the [squares]
You need to pay more attention to what is given & asked for.
x + (x + 2)+ (x + 4)= 116 [x^2+(x+2)^2+(x+4)^2=116]
3x + 6 = 116 [x^2+x^2+4x+4+x^2+8x+16=116]
-6 - 6 [3x^2+12x+20-116=0]
[3x^2 +12x-96=0]
3x = 110 [3(x^2+4x-32)=0]
3x/3 110/3 3(x+8)(x-4)=0]
x = 36.6666 [x-4=0]
[x=4]
Proof:
4^2+6^2+8^2=116
16+36+64=116
116=116

SOLUTION: The sum of the squares of three consecutive even integers is 116. Find the integers... This is what I have tried... x + (x + 2)+ (x + 4)= 116 3x + 6 = 116 -6 - 6 3x =