the age of a man is twice the square of the ages of his sons . eight years hence the age of the man will be 4 years more than three time the age ohf his son find their present ages.
Answers
ANSWER:-
Given:
The age of a man is twice the square of the ages of his sons. 8 years hence the age of the man will be 4 years more than 3 times the age of his son.
To find:
Find their present ages.
Solution:
Let the present age of his son be x years.&
Let the present age of the father be 2x² years.
After 8 years:
Age of son= (x+8) years
Age of father=(2x² +8) years.
According to the question:
=) 2x² +8 = 3(x+8)+4
=) 2x² +8 = 3x +24 +4
=) 2x² + 8 = 3x +28
=) 2x² -3x +8 -28=0
=) 2x² -3x -20 =0
=) 2x² -8x +5x -20 =0
=) 2x(x-4) +5(x-4) = 0
=) (x-4)(2x+5)=0
=) x-4 =0 or 2x+5=0
=) x= 4 or x= -5/2
Since, age can't be negative.
So,
x = 4
Age of son =x = 4 years.
Age of father = 2x² =2(4)²
=) 2× 16
=) 32 years
Hope it helps ☺️
First take man's age = x then second case son's age = x
Solution:
Let the present age of son be years
Present age of his father years
Age of son after 8 years years
Age of his father after 8 years years
Given age of man will be 4 years more than 3 times his sons age.
Hence
2x^2+8=3x+28
2x^2-3x-20=0
2x^2-8x+5x-20=0
2x(x-4)+5(x-4)=0
(x-4)(2x+5)=0
x=4;x=-5/2
therefore x=4
therefore present age of son is 4 years
therefore present age of his father age 32 years