Question

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.​

Answer 1

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 ☺️

Answer 2

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