 ## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 51 of Exercise 10A

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51
##### 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 1
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