Book: RS Aggarwal - Mathematics
Chapter: 10. Quadratic Equations
Subject: Maths - Class 10th
Q. No. 10 of Exercise 10E
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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
minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.
