Question

the age of a man is twice the square of the age of his son eight years hence, the age of the man will be 4 years more than three times the age of his son. find their present ages​

Answer 1

Answer

Let the present age of son be x years.

So, the present age of man is (2x²) years

Age of son 8 years hence = (x + 8) years

Age of man 8 years hence = (2x² + 8) years

So, $\sf ({2x}^{2} + 8) = 3 (x + 8) + 4$

$\sf \dashrightarrow {2x}^{2} - 3x - 30 = 0$

$\sf \dashrightarrow {2x}^{2} - 8x + 5x - 20 = 0$

$\sf \dashrightarrow 2x (x - 4) + 5 (x - 4) = 0$

$\sf \dashrightarrow (x - 4) (2x + 5) = 0$

$\sf \dashrightarrow x - 4 = 0 \: \: or \: \: 2x + 5 = 0$

$\sf \dashrightarrow x = 4 \: \: or \: \: x = \dfrac{-5}{2}$

Age cannot be negative. So,

$\sf \dashrightarrow x = 4$

So,

$\sf Son's \: present \: age = 4 \: years$

$\sf \dashrightarrow (2 \times {4}^{2}) \: years$

$\sf Man's \: present \: age = 32 \: years$

Hence, the age of son is 4 years and the age of man is 32 years.