The age of a man is twice the square of the age of his son. Eight year

1.	The age of a man is twice the square of the age of his son. Eight yearshence, the age of the man will be 4 years more than three times theage of his son. Find their present ages.ICBSE 2009C
Answer» Let the sons'age be = x father's age = 2x^2 8 years later... son = x+8 Father = 2 x^2 +8 Since the difference between father and son age is 4. According to question, 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 (2x-5)(x+4) = 0 x+4 = 0 x = -4 ( age cannot be negative so we leave out the negative sign ) so son's age is 4. father age = 2x^2 = 32 answer

The age of a man is twice the square of the age of his son. Eight yearshence, the age of the man will be 4 years more than three times theage of his son. Find their present ages.ICBSE 2009C

Answer»

Let the sons'age be = x

father's age = 2x^2

8 years later...

son = x+8

Father = 2 x^2 +8

Since the difference between father and son age is 4.

According to question,

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

(2x-5)(x+4) = 0

x+4 = 0

x = -4 ( age cannot be negative so we leave out the negative sign )

so son's age is 4.

father age = 2x^2 = 32 answer