Question

Two years ago a man's age was 3 times the square of his son's age. In 3 years time, his age will be four times his son's age. Find there present age.

Answer 1

at present
son age = x years
father age =y
two years ago
son age =(x-2)
father age =(y-2)
father age=3×(sonage)^2
y-2=3(x-2)^2-----(1)
after 3 years
son age= (x+3)
father age=(y+3)
father age=4×son age
y+3=4(x+3)
y=4x+12-3
y=4x+9---(2)
substitute (2) in (1)
4x+9-2=3(x-2)^2
4x+7=3x^2-12x+12
0=3x^2-16x+5
3x^2-16x+5=0
3x^2-x-15x+5=0
x(3x-1)-5(3x-1)=0
(3x-1)(x-5)=0
therefore
3x-1=0 or x=5
we take x=5
substitute x=5 in (2)
y=4×5+9
y=29
son age=x=5 years
father age=y=29 years