Quadratic equation. Class 10th maths
Dear student
Let the present age of son be x years.
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(x 2 – 4x + 4)+ 5 = 4x + 12
⇒ 3x 2 – 16x+ 5 = 0
⇒ 3x 2 – 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.
Hope! This will help you.
Cheers!