Question

Two years ago, a man’s age was three times the square of his daughter’s age. Three years hence, his age will be four times his daughter’s age. Find their present ages.

Answer 1

2 years ago,
Let the age of daughter = x
age of man = 3x²
then present age of daughter = x + 2
and mean = 3x² + 2
and 3 years hence , the age of
the daughter = x + 2 + 3 = x + 5
and man = 3x² + 2 + 3 = 3x² + 5
According to the condition.
3x² + 5 = 4(x + 5)
⇒ 3x² + 5 = 4x + 20
⇒ 3x² - 4x + 5 - 20 = 0
⇒ 3x² - 4x - 15 = 0
⇒ 3x² - 9x + 5x - 15 = 0
⇒ 3x(x - 3) + 5(x - 3) = 0
⇒ (x - 3)(3x + 5) = 0
EIther x - 3 = 0,
then x = 3
or
3x + 5 = 0, 
then 3x = -5

⇒ x = `(-5)/(3)`
Which is not possible, as age can't be negative
If x = 3, then
Present age of man
= 3x² + 2
= 3(3)² + 2
= 27 + 2
= 29 years
and age of daughter
= x + 2
= 3 + 2 
= 5 years.