# 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 Ax2 + Bx + C is divided in two such values that:

Product = AC

For the given equation A = 9, B = - 9(a + b), C = 2a2 + 5ab + 2b2

= 9(2a2 + 5ab + 2b2)
= 9(2a2 + 4ab + ab + 2b2)
= 9[2a(a + 2b) + b(a + 2b)]
= 9(a + 2b)(2a +  b)
= 3(a + 2b)3(2a + b)

Also,
3(a + 2b) + 3(2a + b) = 9(a + b)

Therefore,

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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