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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.

Respuesta :

jcherry99
Let the middle child be x
youngest child = x - 1
Oldest child = x + 1

(x - 1)^2 = 8(x + 1) + 4
x^2 - 2x + 1 = 8x + 8 + 4 
x^2 - 10x - 11 = 0
(x - 11) * (x + 1) = 0

X = - 1 makes no sense. How can a middle child be - 1 years old?
x = 11

The youngest child is 10
The middle child is 11
The oldest child is 12

Check
=====
10^2 = 100
(8*12) + 4
96 + 4 = 100 So the results have been checked.

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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