Question

Solve the following quadratic equation:

9x²-9(a+b)x+[2a²+5ab+2b²]=0.

Answer 1

9x² - 9(a + b)x + (2a² + 4ab + ab + ab²) = 0

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

9x² - (9a + 9b)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)][3x - (2a + b)] = 0

[3x - (a + 2b)] = 0 

or we can have

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

3x = (a + 2b) or 3x = (2a + b)

Hence, x = \(\frac{a+2b}{3}\) or x = \(\frac{2a+b}{3}\)

Answer 2

Answer is....

Comments

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Answer 3

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

⇒ 3x = a + 2b

⇒ x = \(\frac{a + 2b}3\)

or,

⇒ 3x = 2a + b

⇒ x = \(\frac{2b+b}3\)