#### 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.

#### Solution

Let the present age of the son be x years.

∴ Present age of father =` 2x^2 `years

Eight years hence,

Son's age = (x + 8) years

Father's age =` (2x^2 + 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.

`∴ 2x^2 + 8 = 3(x + 8) + 4`

`2x^2 + 8 = 3x + 24 + 4`

`2x^2 − 3x − 20 = 0`

`2x^2 − 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.

Is there an error in this question or solution?

Solution 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. Concept: Quadratic Equations.