Two natural numbers differ by 3 and their product is 504. Find the numbers.

Let the two natural numbers be x and (x + 3)

According to given condition,


x(x + 3) = 504


x2 + 3x – 504 = 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 = 3 c = – 504


= 1. – 504 = – 504


And either of their sum or difference = b


= 3


Thus the two terms are 24 and – 21


Difference = 24 – 21 = 3


Product = 24. – 21 = – 504


x2 + 3x – 504 = 0


x2 + 24x – 21x – 504 = 0


x (x + 24) – 21(x + 24) = 0


(x + 24) (x – 21) = 0


(x + 24) = 0 or (x – 21) = 0


x = – 24 or x = 21


Case I: x = 21


x + 3 = 21 + 3 = 24


The numbers are (21, 24)


Case I: x = – 24


x + 3 = – 24 + 3 = – 21


The numbers are ( – 24, – 21)


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