Question

One year ago, a man was 8 times as old as his son. now, his age is equal to the square of his son's age in years. find their present ages.

Answer 1

☆heya your answer is here☆


Let the present age of son be x years
Therefore, his father's age = x2years

One year ago, there ages would have been (x-1) and (x2 -1) years respectively.
a/c x2 -1/ x-1 =8
x2 -1= 8x- 8
x2- 8x+7=0
x2 - 7x -x+7=0
x (x -7) -1(x-7)=0
(x-1) (x-7)=0
Therefore, x=1 or x=7
Case 1: when x=1
Therefore, his father's age =12= 1 year
It is not possible.
Hence, x=7 is the required solution.
Therefore, son's present age= 7 years
father's present age= 72=49 years