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. Find their present ages.

Let the present age of son be x years

The present age of man = x2 years


One year ago age of son = (x – 1)years


age of man = (x2 – 1)years


According to given question


x2 – 1 = 8(x – 1)


x2 – 1 = 8x – 8


x2 – 8x + 7 = 0


Using the splitting middle term – the middle term of the general equation is divided in two such values that:


Product = a.c


For the given equation a = 1 b = – 8 c = 7


= 1.7 = 7


And either of their sum or difference = b


= – 8


Thus the two terms are – 7 and – 1


Sum = – 7 – 1 = – 8


Product = – 7. – 1 = 7


x2 – 7x – x + 7 = 0


x(x – 7) – 1(x – 7) = 0


(x – 7) (x – 1) = 0


x = 1 or x = 7


Man’s age cannot be 1 year


Thus x = 7


Thus the present age of son is 7years


The present age of man is 72 = 49years


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