Book: RS Aggarwal - Mathematics
Chapter: 10. Quadratic Equations
Subject: Maths - Class 10th
Q. No. 51 of Exercise 10A
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Solve each of the following quadratic equations:
9x2 - 9 (a + b)x + (2 a2 + 5ab + 2b2) = 0
9x2 - 9(a + b)x + (2 a2 + 5ab + 2b2) = 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 = 9 b = - 9(a + b) c = 2a2 + 5ab + 2b2
= 9(2a2 + 5ab + 2b2)
And either of their sum or difference = b
= - 9(a + b)
Thus the two terms are - 3(2a + b) and - 3(a + 2b)
Sum = - 3(2a + b) - 3(a + 2b)
= - 6a - 3b - 3a - 6b
= - 9a - 9b
= - 9(a + b)
Product = - 3(2a + b). - 3(a + 2b)
= 9(2a + b)(a + 2b)
= 9(2a2 + 5ab + 2b2)
9x2 - 9 (a + b) x + (2 a2 + 5ab + 2b2) = 0
9x2 - 3(2a + b)x - 3(a + 2b)x + (a + 2b) (2a + b) = 0
3x[3x - (2a + b)] - (a + 2b)[3x - (2a + b)] = 0
[3x - (2a + b)] [3x - (a + 2b)] = 0
[3x - (a + 2b)] = 0 or [3x - (2a + b)] = 0
Hence the roots of equation are 
