MATH SOLVE

3 months ago

Q:
# 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.

Accepted Solution

A:

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 + 4Β

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.

youngest child = x - 1

Oldest child = x + 1

(x - 1)^2 = 8(x + 1) + 4

x^2 - 2x + 1 = 8x + 8 + 4Β

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.