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 = x^{2} years

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

age of man = (x^{2} – 1)years

According to given question

x^{2} – 1 = 8(x – 1)

x^{2} – 1 = 8x – 8

x^{2} – 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

x^{2} – 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 7^{2} = 49years

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