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.

Answer: 10, 11, & 12

Step-by-step explanation:

Let x represent the age of the youngest child.

Their ages are consecutive so,

Youngest: x

Middle: x + 1

Oldest: x + 2

The age of the Youngest squared (x²) equals 8 times the Oldest [8 (x + 2) ] plus 4.

x² = 8 (x + 2) + 4

x² = 8x + 16 + 4

x² = 8x + 20

x² - 8x - 20 = 0

(x - 10) (x + 2) = 0

x - 10 = 0 or x + 2 = 0

x = 10 or x = - 2

Since age cannot be negative, x = - 2 is not valid

So, the Youngest (x) is 10

the Middle (x + 1) is 11

and the Oldest (x + 2) is 12