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


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