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.

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

225x²+4=0

225x²=0-4

225x²=-4

x²=-4/225

x=±√-4/225