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Question
Paul is x years old and his father’s age is twice the square of Paul’s age. Ten years hence, the father’s age will be four times Paul’s age. Find their present ages.
Solution
Age of Paul = x years
Father's age = 2x2
10 years hence,
Age of Paul = x + 10
and father's age = 2x2 + 10
According to the conditions,
2x2 + 10 = 4(x + 10)
⇒ 2x2 + 10 = 4x + 40
⇒ 2x2 + 10 - 4x - 40 = 0
⇒ 2x2 - 4x - 30 = 0
⇒ x2 - 2x - 15 = 0 ...(Dividing by 2)
⇒ x2 - 5x + 3x - 15 = 0
⇒ x(x - 5) + 3(x - 5) = 0
⇒ (x - 5)(x + 3) = 0
Either x - 5 = 0,
then x = 5
or
x + 3 = 0,
then x = -3,
but it is not possible as it is in negative.
∴ Age of Paul = 5 years.
and his father's age
= 2x2
= 2(5)2
= 2 x 25
= 50 years.
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