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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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Answer

Let present age of son be x years

Two years before age of,

Son = (x - 2) years

Man = 3(x - 2)2 years

Man's present age = 3(x - 2)2 + 2

In three years,

Son = x + 3 years

Man = 4(x + 3) years

We can write,

⇒ 3(x - 2)2 + 5 = 4(x + 3)

⇒ 3(x2 + 4 - 4x) + 5 = 4x + 12

⇒ 3x2 - 12x + 12 + 5 = 4x + 12

⇒ 3x2 - 12x - 4x + 17 - 12 = 0

⇒ 3x2 - 16x + 5 = 0

⇒ 3x2 - 15x - x + 5 = 0

⇒ 3x(x - 5) - 1(x - 5) = 0

⇒ (3x - 1)(x - 5) = 0

⇒ 3x - 1 = 0 or x - 5 = 0

⇒ x = 13\dfrac{1}{3} or x = 5.

x ≠ 13\dfrac{1}{3} as in this case (x - 2) will be negative and age cannot be negative.

Man's present age = 3(x - 2)2 + 2 = 3(5 - 2)2 + 2 = 3(3)2 + 2 = 29 years.

Hence, son's present age = 5 years and man's present age = 29 years.

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