Question

Solve 9x2-9(a+b)x+(2a2+5ab+2b2)=0 by factorisation

Answer 1

Solve 9x2-9(a+b)x+(2a2+5ab+2b2)=0 by factorisation
   Consider, 9x2-9(a+b)x+(2a2+5ab+2b2)=0

9x2 - 9(a + b)x + (2a2 + 4ab + ab + 2b2) = 0

9x2 - 9(a + b)x + [2a(a + 2b) + b(a + 2b)] = 0

9x2 - 9(a + b)x + [(a + 2b)(2a + b)] = 0

9x2 – 3[(a + 2b) + (2a + b)]x + [(a + 2b)(2a + b)] = 0

9x2 – 3(a + 2b)x - 3(2a + b)x + [(a + 2b)(2a + b)] = 0

3x[3x  – (a + 2b)] - (2a + b)[3x + (a – 2b)] = 0

[3x  – (a + 2b)][3x - (2a + b)] = 0

[3x – (a + 2b)] = 0 or [3x - (2a + b)] = 0

3x  =  (a + 2b) or 3x = (2a + b)
hope it helps
x = (a + 2b)/3 or x = (2a + b)/3    

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Here we have,

9x² - 9(a + b)x + (2a² + 5ab + 2b²) = 0

So,

= 2a² + 5ab + 2b² (This is a constant term)

Splitting middle term :-

= 2a² + 4ab + ab + 2b²

= 2a(a + 2b) + b(a + 2b)

= (a + 2b) (2a + b)

Now,

Coefficient of middle term

= -9(a + b)

= -3[(2a + b) + (a + 2b)]

So,

9x² - 9(a + b)x + (2a² + 5ab + 2b²) = 0

9x² - 3[(2a + b) + (a + 2b)]x + (2a + b) (a + 2b) = 0

9x² - 3(2a + b)x - 3(a + 2b)x + (2a + b) (a + 2b) = 0

3x[3x - (2a + b) - (a + 2b)][3x - (2a + b)] = 0

[3x - (2a + b)][3x - (a + 2b)] = 0

[3x - (2a + b)] = 0

x = 2a + b/3

[3x - (a + 2b)] = 0

x = a + 2b/3