Question

The age of a father is twice the square of the age of his son. Eight years hence, the age of the father will be 4 years more than three times the age of the son. Find their present ages.

Answer 1

Let the present age of the son be x years.
∴ Present age of father =` 2x^2 `years
Eight years hence,
Son's age = (x + 8) years
Father's age =` (2x^2 + 8)` years
It is given that eight years hence, the age of the father will be 4 years more than three  times the age of the son. 

`∴ 2x^2 + 8 = 3(x + 8) + 4`
`2x^2 + 8 = 3x + 24 + 4`
`2x^2 − 3x − 20 = 0`
`2x^2 − 8x + 5x − 20 = 0`
`2x(x − 4) + 5(x − 4) = 0`
`(x - 4) (2x + 5) = 0` 

`x = 4, (-5)/2` 

But, the age cannot be negative, so, x = 4.
Present age of son = 4 years
Present age of father = 2(4)² years = 32 years.