Question

Six years before, the age of the mother was equal to the square of her son's age. Three years hence, her age will be thrice the age of her son then find the present ages of the mother and son.

Answer 1

Let age of son $6$ years before be $x$

then mother's age = $x^2$

then, persent age son $=x+6$

mothers age $=x^2+6$

After $3$ year,

son $=x+6+3=x+9$

mother $=x^2+6+3=x^2+9$

A.T.Q

$x^2+9=3(x+9)$

$x^2+9=3x+27$

$x^2-3x-18=0$

$x^2-6x+3x-18=0$

$x(x-6)+3(x-6)=0$

$(x+3)(x-6)=0$

$x=6$ (-3 will neglect)

Persent age of mother = $x^2+6\Rightarrow 36+6=42$

Son = $x+6=6+6=12$