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 = 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

Concept: Quadratic Equations

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