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.
Let son's age 2 years ago be x years. Then,
Man's age 2 years ago = 3x2 years
∴ Son's present age = (x + 2) years
Man's present age = (3x2 + 2) years
In three years time,
Son's age = (x + 2 + 3) years = (x + 5) years
Man's age = (3x2 + 2 + 3) years = (3x2 + 5) years
According to the given condition,
Man's age = 4 × Son's age
∴ 3x2 + 5 = 4(x + 5)
$\Rightarrow 3 x^{2}+5=4 x+20$
$\Rightarrow 3 x^{2}-4 x-15=0$
$\Rightarrow 3 x^{2}-9 x+5 x-15=0$
$\Rightarrow 3 x(x-3)+5(x-3)=0$
$\Rightarrow(x-3)(3 x+5)=0$
$\Rightarrow x-3=0$ or $3 x+5=0$
$\Rightarrow x=3$ or $x=-\frac{5}{3}$
∴ x = 3 (Age cannot be negative)
Son's present age = (x + 2) years = (3 + 2) years = 5 years
Man's present age = (3x2 + 2) years = (3 × 9 + 2) years = 29 years