SOLUTION: david`s father is two year old than his mother and his mother`s age is the square of his age. They years adds up to 80 years.How old is each one of them?

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Question 5955: david`s father is two year old than his mother and his mother`s age is the square of his age. They years adds up to 80 years.How old is each one of them?
Answer by glabow(165) About Me  (Show Source):
You can put this solution on YOUR website!
Define variables:
x = David's age
f = father's age
m = mother's age
We know:
f = m + 2
x%5E2 = m
x + f + m = 80
Rearrange to x+=+80+-+f+-+m
Substitute m+2 for f, giving x=80-%28m%2B2%29-m which is x=80-m-2-m
which is x=80-2m-2 which is x=78-2m
Substitute x%5E2 for m, giving x=78-2x%5E2
Rearranging gives 2x%5E2%2Bx-78=0
This can be solved by factoring. We are looking for two numbers that multiply to -78 and when we subtract twice one from the other we get 1. If we assume the ages are given in whole numbers, possible answers to numbers that multiply to 78 are (1, 78), (2, 39), and (6, 13). By inspection we see that 2*6 is different from 13 by 1. So the factors are either 6 and -13 or -6 and 13. Since we want a positive 1, -6 and 13 are correct. The equation factors to
x=2x%5E2%2Bx-78=%282x%2B13%29%28x-6%29=0 and x will be either -13 or 6.
Obviously,
x = 6
m = 36
f = 38
Checking, 6%2B36%2B38=80