Book: RS Aggarwal - Mathematics
Chapter: 10. Quadratic Equations
Subject: Maths - Class 10th
Q. No. 9 of Exercise 10E
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Find two consecutive multiples of 3 whose product is 648.
Let the required consecutive multiples of 3 be 3x and 3(x + 1)
According to given condition,
3x.3(x + 1) = 648
9(x2 + x) = 648
x2 + x = 72
x2 + x – 72 = 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 = 1 c = – 72
= 1. – 72 = – 72
And either of their sum or difference = b
= 1
Thus the two terms are 9 and – 8
Difference = 9 – 8 = 1
Product = 9. – 8 = – 72
x2 + 9x – 8x – 72 = 0
x (x + 9) – 8(x + 9) = 0
(x + 9) (x – 8) = 0
(x + 9) = 0 or (x – 8) = 0
x = – 9 or x = 8
x = 8 (rejecting the negative values)
3x = 3.8 = 24
3(x + 1) = 3(8 + 9) = 3.9 = 27
Hence, the required numbers are 24 and 27
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.