Question

Two years ago a man’s age was three times the square of his son’s age. In three yrs time, his age Will be 4 times his son’s age. Find their present ages.​

Answer 1

Answer:

Let the age of son 2 years ago be x years

. Then, father's age 2 years ago = 3x2 years

Present age of son = (x + 2) years

Present age of father = (3x2 + 2) years

Answer 2

Let the present age of son = x years.

Now according to your question

Before two years age of son = (x - 2) years.

age of father before two years = 3(x - 2)²

years.

Then,

present age of father {3(x - 2)² + 2} years

again according to question

After three years;

age of son will be (x + 3) year

and

age of father will be {3(x - 2)² + 2+ 3} years = {3(x − 2)² + 5}years -

again question say that

3(x - 2)² + 5 = 4(x + 3)

=> 3(x² - 4x + 4)+ 5 = 4x + 12

=> 3x² - 16x+ 5 = 0

=> 3x² - 15x-x+ 5 = 0

=> 3x²- 15x-x+ 5 = 0

=> 3x (x - 5)-1×(x-5) = 0

=> (3x − 1) (x − 5) = 0 - -

=> 3x - 1 = 0 or x − 5 = 0

=> x = 1/3 or 5

If x = 1/3, then age will be negative So it is not possible

and

If x = 5, the present age of son is 5 years

and

hence present age of father is= (3 × 32 + 2) = 29 years.

Hope it helps....