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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