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.
2006-11-14 15:50:38 · 6 answers · asked by <3 2 in Science & Mathematics ➔ Mathematics
THE OLDEST ONE IS 12, THE YOUEST 10. SO THE MIDDLE MUST BE 11
2006-11-14 15:55:01 · answer #1 · answered by Mike P 3 · 0⤊ 1⤋
Let us begin by assigning the variable x to represent the age of the youngest child. Therefore, the ages of the other two children can be represented as (x + 1) and (x + 2).
If the square of the age of the youngest child is 4 more then 8 times the age of the oldest child, we can represent it with the following equation...
(x^2) = 8 (x + 2) + 4
...which we can then reduce...
x^2 = 8 (x + 2) + 4
x^2 = 8x + 16 + 4
x^2 = 8x + 20
x^2 - 8x - 20 = 0
Using the quadratic equation, we solve for x.
x = -2 and 10.
Since the age of a child cannot be negative, we know the age of the youngest child is 10 years. Thusly, the ages of the other two children are 11 and 12 years.
2006-11-14 15:59:15 · answer #2 · answered by An N 2 · 0⤊ 0⤋
youngest = 10 yrs
middle = 11 yrs
oldest = 12 yrs
2006-11-14 15:54:37 · answer #3 · answered by jacobrcotton 3 · 0⤊ 0⤋
youngest=10 yrs
middle one = 11 yrs
oldest=12 yrs.
2006-11-15 03:17:57 · answer #4 · answered by virendra s 2 · 0⤊ 0⤋
let age of the youngest be x then all three ages are : x,x+1,x+2
also x^2-8(x+2) = 4
x^2-8x-20 = 0
x^2-10x+2x-20=0
x(x-10) +2(x-10)=0
(x+2)(x-10)=0
so x = 10 or -2
disregarding -2 x=10
so ages are x=10,11,12
2006-11-14 16:00:39 · answer #5 · answered by yog 2 · 0⤊ 0⤋
8
16
24
32
2006-11-14 15:54:45 · answer #6 · answered by gussie r 3 · 0⤊ 0⤋