Paul is x years old and his father's age is twice the square of Paul's age. Ten years hence, father's age will be four times Paul's age. Find their present ages.

Let paul's present age be x years

Father's age = 2x²

After 10 years,

Paul's age = (x + 10) years

Father's age = (2x² + 10)

According to question,

$\Rightarrow 2x^2 + 10 = 4(x + 10) \\[1em] \Rightarrow 2x^2 + 10 = 4x + 40 \\[1em] \Rightarrow 2x^2 - 4x + 10 - 40 = 0 \\[1em] \Rightarrow 2x^2 - 4x - 30 = 0 \\[1em] \Rightarrow 2(x^2 - 2x - 15)= 0 \\[1em] \Rightarrow x^2 - 2x - 15 = 0 \\[1em] \Rightarrow x^2 - 5x + 3x - 15 = 0 \\[1em] \Rightarrow x(x - 5) + 3( x - 5) = 0 \\[1em] \Rightarrow (x + 3)(x - 5) = 0\\[1em] \Rightarrow x + 3 = 0 \text{ or } x - 5 = 0 \\[1em] x = -3 \text{ or } x = 5$

Since, age cannot be negative hence x ≠ -3

∴ x = 5 , 2x² = 50

Paul's age is 5 years , while his father's age is 50 years.