SOLUTION: A man is 3 times as old as his son. 8 years ago the product of their age was 112. Find their ages

Question 519251: A man is 3 times as old as his son. 8 years ago the product of their age was 112. Find their ages
Found 2 solutions by mananth, Maths68:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
son = x
father = 3x
8 years ago
son = x-8
father = 3x-8
(x-8)(3x-8)=112
3x^2-32x+64=112
3x^2-32x-48=0
Find the roots of the equation by quadratic formula

a= 3 ,b= -32 ,c= -48

b^2-4ac= 1024 + 576
b^2-4ac= 1600

x1=( 32 + 40 )/ 6
x1= 12
x2=( 32 -40 ) / 6
x2= -1.33
Ignore negative value
son = 12 yaers
father = 36 years
m.ananth@hotmail.ca

Answer by Maths68(1474) (Show Source): You can put this solution on YOUR website!
Let
Present age of Man = m
Present age of Son = s

A man is 3 times as old as his son.
m=3s.......(1)
8 years ago,
Man age = m-8
Song age = s-8
Then, the product of their age was 112.
(m-8)(s-8)=112
ms-m8-8s+64=112
ms-m8-8s=112-64
ms-m8-8s=48..............(2)
Substitute the value from (1) to (2)
(3s)s-(3s)8-8s=48
3s^2-24s-8s=48
3s^2-32s=48
3s^2-32s-48=0
3s^2-36s+4s-48=0
3s(s-12)+4(s-12)=0
(s-12)(3s+4)=0
s-12=0 or 3s+4=0
s=12 or s=-4/3 (inadmissable)
so
s=12
Put the value of s in (1)
m=3s
m=3(12)
m=36
Present age of Man = m = 36 years old
Present age of Son = s = 12 years old

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