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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. - Mathematics

Numerical
Sum

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.

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Solution

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

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