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 the son be x years.

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

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

Age of the man 8 years hence (2x² + 8) years.

$\sf \therefore \: (2x² + 8) = 3(x+8) + 4$

$\sf \longrightarrow 2x²-3x-20 = 0$

$\sf \longrightarrow 2x²-8x+5x-20 = 0$

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

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

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

$\sf \longrightarrow x = 4 \: \: or \: \: x = \frac{ - 5}{2}$

$\sf \longrightarrow x = 4$

ㅤㅤㅤㅤ[ because age cannot be negative ]

son's present age = 4 years, and

man's present age = (2 x 4²) years = 32 years.