Question

06. Two years ago, a man's age was three times the square of his son's age. in three years, his age
will be four times his son's age. Find their present ages.​

Answer 1

Refer to the attached image ❤️

Answer 2

2 Years ago……

Age of Father = (F - 2) years

Age of Son = (S - 2) years

Therefore,

(F - 2) = 3 x (S - 2)^2

(F - 2) = 3 x (S^2 + 4 - 4S)

(F - 2) = 3xS^2 + 12 - 12S

3xS^2 - F - 12S + 14 = 0

3xS^2 -12S + 14 = F —————————Equation no. 1

Three years hence, his age will be 4 times his son's age

3 Years hence……

Age of Father = (F + 3) years

Age of Son = (S + 3) years

Therefore,

(F + 3) = 4 x (S + 3)

(F + 3) = 4xS + 12

4S - F + 9 = 0

4S + 9 = F —————————————Equation no. 2

Equating equation 1 and 2,

3xS^2 -12S + 14 = 4S + 9

3xS^2 - 16S + 5 = 0

Solving this equation,

S= 1/3 or 5.

As age cannot be a fraction,

therefore, S = 5.

Therefore F = 29