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.
Quadratic Equations
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Question

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.

KnowledgeBoat · Accepted Answer

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 = or y = 5 Since, 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.

Mathematics

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.

Quadratic Equations

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