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
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answer is 20726 it was easy man but equation was nice
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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..
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..
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