RS Aggarwal - Mathematics

Book: RS Aggarwal - Mathematics

Chapter: 10. Quadratic Equations

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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