#### 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.

#### Solution

Let the age of son 2 years ago be x years.

Then, father’s age 2 years ago = 3x^{2} years

Present age of son = (x + 2) years

Present age of father = (3x^{2} + 2) years

3 years hence:

Son’s age = (x + 2 + 3) years = (x + 5) years

Father’s age = (3x^{2} + 2 + 3) years = (3x^{2} + 5) years

From the given information,

3x^{2} + 5 = 4(x + 5)

3x^{2} – 4x – 15 = 0

3x^{2} – 9x + 5x – 15 = 0

3x(x – 3) + 5(x – 3) = 0

(x – 3) (3x + 5) = 0

x = 3,

Since, age cannot be negative. So, x = 3.

Present age of son = (x + 2) years = 5 years

Present age of father = (3x^{2} + 2) years = 29 years

Is there an error in this question or solution?

Solution 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. Concept: Quadratic Equations.