Two years ago, a man’s age was three times the square of his son’s age. In three years time, his age will be four times his son’s age. Find their present ages.

Let the present age of man be x and his son be y.

Given,

Two years ago, man’s age was three times the square of son’s age.

⇒ x - 2 = 3(y - 2)²     ………(1)

Given,

In three years time, man’s age will be four times son’s age.

⇒ x + 3 = 4(y + 3)

⇒ x + 3 = 4y + 12

⇒ x = 4y + 12 - 3

⇒ x = 4y + 9     ………(2)

Substituting value of x from equation (2) in equation (1), we get :

⇒ 4y + 9 - 2 = 3(y - 2)²

⇒ 4y + 9 - 2 = 3(y² - 4y + 4 )

⇒ 4y + 7 = 3y² - 12y + 12

⇒ 3y² - 12y + 12 - 4y - 7 = 0

⇒ 3y² - 16y + 5 = 0

⇒ 3y² - 15y - y + 5 = 0

⇒ 3y(y - 5) - 1(y - 5) = 0

⇒ (3y - 1)(y - 5) = 0

⇒ (3y - 1) = 0 or (y - 5) = 0     [Using zero - product rule]

⇒ y = $\dfrac{1}{3}$ or y = 5

Since, $\dfrac{1}{3}$ is not whole number, thus y = 5.

Substituting value of y in equation (2), we get :

⇒ x = 4(5) + 9 = 20 + 9 = 29.

Hence, age of man = 29 years and age of his son = 5 years.