## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

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

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10
##### Find two consecutive positive odd integers whose product is 483.

Let the required consecutive positive odd integers be x and (x + 2)

According to given condition,

x (x + 2) = 483

x2 + 2x – 483 = 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 = 2 c = – 483

= 1. – 483 = – 483

And either of their sum or difference = b

= 2

Thus the two terms are 23 and – 21

Difference = 23 – 21 = 2

Product = 23. – 21 = – 483

x2 + 2x – 483 = 0

x2 + 23x – 21x – 483 = 0

x (x + 23) – 21(x + 23) = 0

(x + 23) (x – 21) = 0

(x + 23) = 0 or (x – 21) = 0

x = – 23 or x = 21

x = 21 (x is a positive odd integer)

x + 2 = 21 + 2 = 23

Hence, the required integers are 21 and 23

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