#### Question

**The age of the 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 the 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.