Question

Two year ago , a man age was three times the square of his son's age. In three years time, his age will be four. Find their present ages. This chapter is from Quadratic equation. plzzz its urgent I will mark brainlist. plzz

Answer 1

Hey friend!!!

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

=> 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._______answer

hope it will help you..