**Question:**

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.

**Solution:**

Let son's age 2 years ago be *x* years. Then,

Man's age 2 years ago = 3*x*2 years

∴ Son's present age = (*x* + 2) years

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

In three years time,

Son's age = (*x* + 2 + 3) years = (*x* + 5) years

Man's age = (3*x*2 + 2 + 3) years = (3*x*2 + 5) years

According to the given condition,

Man's age = 4 × Son's age

∴ 3*x*2 + 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 = (3*x*2 + 2) years = (3 × 9 + 2) years = 29 years