# Solve each of the

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 Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 