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Question
Solve the following quadratic equation:
9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 0
Sum
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Solution
We write, –9(a + b)x = –3(2a + b)x – 3(a + 2b)x as
9x2 × (2a2 + 5ab + 2b2) = 9(2a2 + 5ab + 2b2)x2 = [–3(2a + b)x] × [–3(a + 2b)x]
∴ 9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 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` or `x = (a + 2b)/3`
Hence, `(2a + b)/3` and `(a + 2b)/3` are the roots of the given equation.
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