Question

Two years ago,a man's age was three times the square of his son's age. In three years's time his age will be four times his son's age. Find their present ages.

Answer 1

Let the man's current age be $x$ and that of his son be $y$

From first condition,

$(x-2)=3(y-2{)}^2$. ----------$\left(1\right)$

From the second condition,

$(x+3)=4(y+3)$

$x+3=4y+12$

$x=4y+9$ ------------$\left(2\right)$

put the value of $x$ from equation $2$ to $1$

$4y+9-2=3(y^2+4-4y)$

$3y^2-16y+5=0$

$3y^2-15y-y+5=0$

$3y(y-5)-1(y-5)=0$

$(3y-1)(y-5)=0$

so, $y=\frac{1}{3}$. which is not possible

and $y=5$

son's age $x=5$ years

father's age $=4\left(5\right)+9=29$ years. (from equation $1$)