Question

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

Answer 1

Let the present age of the son = x years
then, the present age of the man = 2x² years.
8 years hence,
The age of son will be = (x + 8) years and the
age of man = (2x² + 8) years
According to the problem,
2x² + 8 = 3(x + 8) + 4
⇒ 2x² + 8 = 3x + 24 + 4
⇒ 2x² - 3x - 24 - 4 + 8 = 0
⇒ 2x² - 3x - 20 = 0
⇒ 2x² - 8x + 5x - 20 = 0
⇒ 2x(x - 4) +5(x - 4) = 0
⇒ (x - 4)(2x + 5) = 0
EIther x - 4 = 0, 
then x = 4
or
2x + 5 = 0,
then 2x = -5

⇒ x = `-(5)/(2)`
But, it is not possible.
Present age of the son = 4 years
and present age of the man = 2x²
= 2(4)² years
= 32 years.