SOLUTION: The ages of three children in a family can be expressed as consecutive integers. The square of the age of the youngest child is 4 more than 8 times the age of the oldest child. Fin

Question 1130362: The ages of three children in a family can be expressed as consecutive integers. The square of the age of the youngest child is 4 more than 8 times the age of the oldest child. Find the ages of the three children
Found 2 solutions by josgarithmetic, addingup:
Answer by josgarithmetic(39623) (Show Source): You can put this solution on YOUR website!
youngest, n
middle, n+1
oldest, n+2

the description,
Answer by addingup(3677) (Show Source): You can put this solution on YOUR website!
youngest: x
middle : x+1
oldest: x+2
x^2 = 8(x+2)+4
x^2 = 8x + 16 + 4
x^2 = 8x + 20
x^2 - 8x = 20
x^2 - 8x - 20 = 0
(x−10)(x+2)=0
x-10 = 0 or x+2 = 0
x = 10 or x = -2
Let's try the 10:
x^2 = 8(x+2)+4
10^2 = 8(10+2}+4
100 = 80+16+4 Correct, this is your answer
The youngest is 10
The middle is 10+1 = 11
The oldest is 10+2 = 12

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