Two water taps together can fill a tank in 6 hours. The tap of larger diameter takes 9 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Let the time taken by tap of smaller diameter to fill the tank be x hours
The time taken by tap of larger diameter to fill the tank = x – 9 hours
Volume of tank be V
Volume of tank filled by tap of smaller diameter in x hours = V
Volume of tank filled by tap of smaller diameter in 1 hour = V/x
Volume of tank filled by tap of smaller diameter in 6 hours = 
Volume of tank filled by tap of larger diameter in 6 hours = 
Volume of tank filled by tap of smaller diameter in 6 hours + Volume of tank filled by tap of larger diameter in 6 hours = V
12x – 54 = x2 – 9x
x2 – 21x + 54 = 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 = – 21 c = 54
= 1.54 = 54
And either of their sum or difference = b
= – 21
Thus the two terms are – 18 and – 3
Difference = – 18 – 3 = – 21
Product = – 18. – 3 = 54
x2 – 18x – 3x + 54 = 0
x(x – 18) – 3(x – 18) = 0
(x – 18)(x – 3) = 0
(x – 18) = 0 (x – 3) = 0
x = 18 or x = 3
For x = 3 time taken by tap of larger diameter is negative which is not possible
Thus x = 18
Hence the time taken by tap of smaller diameter to fill the tank be 18 hours
The time taken by tap of larger diameter to fill the tank = 18 – 9 = 9hours