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

Let the present age of son be x years.

Before two years age of son was (x – 2) years.

It is given that the age of father before two years = 3(x – 2)2 years.

Thus, the present age of father [3(x – 2)2 + 2] years

After three years; the age of son will be (x + 3) years and the age of father will be [3(x – 2)2 + 2 + 3] years = [3(x – 2)2 + 5] years

 

It is given 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 x – 2 = –5/3.

This means the age of son was –5/3 years before two years. This is impossible since age of a person cannot be negative.

 

If x = 5, the present age of son is 5 years and the present of his father is (3 × 32 + 2) years = 29 years.

Answer 2

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..