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 the present age of the man.
Let the daughter's present age $=x$ years
Then, two years ago, daughter's age $= \left(x-2\right)$ years
$\therefore$ Man's age two years ago $= 3 \left(x-2\right)^2 = 3 \left(x^2 - 4x + 4\right) = 3x^2-12x+12$ years
$\therefore$ Man's present age $= 3x^2 - 12x + 12 + 2 = 3x^2 -12x +14$ years
Man's age three years hence $= 3x^2 - 12x + 14 + 3 = 3x^2 -12x + 17$ years
Daughter's age three years hence $= x+3$ years
$\therefore$ As per sum,
$3x^2-12x+17=4\left(x+3\right)$
i.e. $3x^2-12x+17=4x+12$
i.e. $3x^2-16x+5=0$
i.e. $3x^2 -15x -x + 5 = 0$
i.e. $3x \left(x-5\right)-1 \left(x-5\right)=0$
i.e. $\left(3x-1\right) \left(x-5\right)=0$
i.e. $x=\dfrac{1}{3}$ or $x=5$
If $x=\dfrac{1}{3}$, then man's present age $=3 \times \left(\dfrac{1}{3}\right)^2 - 12 \times \dfrac{1}{3}+14= 10\dfrac{1}{3}$ years, which is not a reasonable value.
$\therefore$ Daughter's present age $=5$ years
$\therefore$ Man's present age $= 3 \times 5^2 - 12 \times 5 + 14 = 29$ years