Question

Two years ago my age was 4$\frac{1}{2}$ times the age of my son .Six years ago, my age was twice the square of the age of my son .What is the present age of my son ?

Answer 1

Let the present age of my son be x years. Then, his age two years ago was (x – 2) years and six years ago, (x – 6) years.
Thus, my age two years ago was $4\frac{1}{2}\left(x-2\right)=\frac{9}{2}\left(x-2\right)$ years and,
six years ago, 2(x – 6)² years.
Thus, we get,
$\frac{9}{2}\left(x-2\right)-2{\left(x-6\right)}^2=4$
$\Rightarrow$ 9(x – 2) – 4(x – 6)² = 8
$\Rightarrow$9x – 18 – 4(x² – 12x + 36) = 8
$\Rightarrow$ 9x – 18 – 4x² + 48x – 144 = 8
$\Rightarrow$ 4x² – 57x + 170 = 0
On splitting the middle term –57x as –40x – 17x, we get:
4x² – 40x – 17x + 170 = 0
$\Rightarrow$ 4x(x – 10) – 17(x – 10) = 0
$\Rightarrow$ (x – 10)(4x – 17) = 0
$\Rightarrow$ x – 10 = 0 or 4x – 17 = 0
$\Rightarrow$ x = 10 or x = $\frac{17}{4}$
Since, the age cannot be in fraction,
Thus, the present age of my son is 10 years.