Question

Solve the following quadratic equation by
factorization method
9x2-9(a+b)x+(2a2+5ab+2b2)​

Answer 1

Given Equation,

$\sf \: {9x}^{2} - 9(a + b)x + (2 {a}^{2} + 5ab + 2 {b}^{2} ) = 0$

Firstly, factorise the constant term :

$\sf \: 2 {a}^{2} + 5ab + {2b}^{2} \\ \\ \dashrightarrow \sf \: 2 {a}^{2} + 4ab + ab + 2 {b}^{2} \\ \\ \dashrightarrow \sf \: 2a( {a}^{} + 2b) +b( a + 2 {b}^{} )\\ \\ \dashrightarrow \sf \: (2a + b)(a + 2b)$

Now,

$\longrightarrow \sf \: {9x}^{2} - 9(a + b)x + (2a + b)(a + 2b) = 0 \\ \\ \longrightarrow \sf \: 9{x}^{2} - 3 \{(2a + b) + (a + 2b) \} + (2a + b)(a + 2b) = 0 \\ \\ \longrightarrow \sf \: 9{x}^{2} - 3(2a + b)x - 3 (a + 2b)x + (2a + b)(a + 2b) = 0 \\ \\ \longrightarrow \sf \: 3x \{3x - (2a + b) \} - (a + 2b) \{3x - (2a + b) \} = 0 \\ \\ \longrightarrow \sf \{ 3x - (a + 2b) \}\{3x - (2a + b) \} = 0 \\ \\ \longrightarrow \sf \{ 3x - (a + 2b) \} = 0 \: \: or \: \: \{3x - (2a + b) \} = 0 \\ \\ \longrightarrow \boxed{ \boxed{ \sf \: x = \dfrac{2a + b}{3} \: or \: \dfrac{a + 2b}{3} }}$

Answer 2

9x2−9(a+b)x+(2a2+5ab+2b2) = 0

⟹9x2−9(a+b)x + [2a(a+2b)+b(a+2b)] = 0

⟹ 9x2−9(a+b)x + [(2a+b)(a+2b)] = 0

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

⟹3x [3x−(a+2b)][3x−(2a+b)] = 0

⟹3 x = a +2b

⟹x = 3a+2b

or,

⟹3x = 2 a + b

⟹x = 32a + b