Book: RS Aggarwal - Mathematics
Chapter: 10. Quadratic Equations
Subject: Maths - Class 10th
Q. No. 18 of Exercise 10E
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Divide 57 into two parts whose product is 680.
Let the two consecutive positive even integers be x and (57 – x)
According to given condition,
x (57 – x) = 680
57x – x2 = 680
x2 – 57x – 680 = 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 = – 57 c = – 680
= 1. – 680 = – 680
And either of their sum or difference = b
= – 57
Thus the two terms are – 40 and – 17
Sum = – 40 – 17 = – 57
Product = – 40. – 17 = – 680
x2 – 57x – 680 = 0
x2 – 40x – 17x – 680 = 0
x (x – 40) – 17(x – 40) = 0
(x – 40) (x – 17) = 0
(x – 40) = 0 or (x – 17) = 0
x = 40 or x = 17
When x = 40
57 – x = 57 – 40 = 17
When x = 17
57 – x = 57 – 17 = 40
Hence the required parts are 17 and 40.
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.
