The ages of three family children can be expressed as consecutive integers. the square of the age of the youngest child is 4 more than eight times the age of the oldest child. find the ages of the three children.

The ages of the three children are 10, 11 and 12

Let the youngest child be x-1

Middle child be x

Oldest child be x+1

Therefore, the equation to solve the question will be:

(x-1)² = 8(x+1) + 4

x² - 2x + 1 = 8x + 8 + 4

Collect like terms

x² - 10x - 11 = 0

x² -11x + x - 11 = 0

x(x - 11) + 1(x - 11) = 0

x - 11 = 0

x = 0 + 11 = 11

x + 1 = 11 + 1 = 12

x - 1 = 11 - 1 = 10

Therefore, the ages of the three children are 10, 11 and 12

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