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 = or x = 5.
x ≠ 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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