## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 8 of Exercise 10E

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8
##### 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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