Question

Factorise:
9x^2 - 9(a + b)x + (2a^2 + 5ab + 2b^2) = 0​

Answer 1

Answer:

Step-by-step explanation:

Here, we have

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

Therefore,

Constant term = 2a² + 5ab + 2b²

= 2a² + 4ab + ab + 2b²

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

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

Coefficient of middle term = - 9(a + b) = - 3[(2a + b) + (a + 2b)

Then,

⇒ 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 or [3x - (a + 2b) = 0

⇒ x = 2a + b/3, a + 2b/3

Hence, x = 2a + b/3, a + 2b/3.

Answer 2

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