Question

The age of a father is twice the square of the age of his son. Eight years hence, the age of the father will be 4 years more than three times the age of the son. Find their present ages.

Answer 1

Let the present age of the son be x years.

∴ Present age of father = 2x² years

Eight years hence,

Son’s age = (x + 8) years

Father’s age = (2x² + 8) years

It is given that eight years hence, the age of the father will be 4 years more than three times the age of the son.

∴ 2x² + 8 = 3(x + 8) + 4

2x² + 8 = 3x + 24 + 4

2x² – 3x – 20 = 0

2x² – 8x + 5x – 20 = 0

2x(x – 4) + 5(x – 4) = 0

(x – 4)(2x + 5) = 0

`x = 4, (-5)/2`

But, the age cannot be negative, so, x = 4.

Present age of son = 4 years

Present age of father = 2(4)² years = 32 years.