Book: RS Aggarwal - Mathematics
Chapter: 10. Quadratic Equations
Subject: Maths - Class 10th
Q. No. 44 of Exercise 10A
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Solve each of the following quadratic equations:
x2 - (2b - 1)x + (b2 - b - 20) = 0
x2 - (2b - 1)x + (b2 - b - 20) = 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 = - (2b - 1); c = b2 - b - 20
= 1(b2 - b - 20)
= (b2 - b - 20)
And either of their sum or difference = b
= - (2b - 1)
Thus the two terms are - (b - 5) and - (b + 4)
Sum = - (b - 5) - (b + 4)
= - b + 5 - b - 4
= - 2b + 1
= - (2b - 1)
Product = - (b - 5) - (b + 4)
= (b - 5) (b + 4)
= b2 - b - 20
x2 - (2b - 1)x + (b2 - b - 20) = 0
⇒ x2 - (b - 5)x - (b + 4)x + (b - 5)(b + 4) = 0
⇒ x[x - (b - 5)] - (b + 4)[x - (b - 5)] = 0
⇒ [x - (b - 5)] [x - (b + 4)] = 0
⇒ [x - (b - 5)] = 0 or [x - (b + 4)] = 0
⇒ x = (b - 5) or x = (b + 4)
Hence the roots of equation are (b - 5) or (b + 4)
