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 3 times the age of his son. Find their present ages.

Let son's present age be x years

Man's age = 2x²

After 8 years,

Son's age = (x + 8) years

Man's age = (2x² + 8)

According to question,

$\Rightarrow 2x^2 + 8 = 3(x + 8) + 4 \\[1em] \Rightarrow 2x^2 + 8 = 3x + 24 + 4 \\[1em] \Rightarrow 2x^2 + 8 = 3x + 28 \\[1em] \Rightarrow 2x^2 - 3x + 8 - 28 = 0 \\[1em] \Rightarrow 2x^2 - 3x - 20 = 0 \\[1em] \Rightarrow 2x^2 - 8x + 5x - 20 = 0 \\[1em] \Rightarrow 2x(x - 4) + 5(x - 4) = 0 \\[1em] \Rightarrow (x - 4)(2x + 5) = 0 \\[1em] \Rightarrow x - 4 = 0 \text{ or } 2x + 5 = 0 \\[1em] x = 4 \text{ or } x = -\dfrac{5}{2}$

Since, age cannot be negative hence x ≠ $-\dfrac{5}{2}$

∴ x = 4 , 2x² = 32

The present age of son is 4 years while the age of man is 32 years.