Using quadratic formula solve the following quadratic equation: [3 MARKS]

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

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Solution

Formula: 1 Mark
Steps: 1 Mark
Answer: 1 Mark

We have,

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

Comparing this equation with $A x^{2} + B x + C = 0$ , we have

$A = 9, B = - 9 (a + b) a n d C = 2 a^{2} + 5 a b + 2 b^{2}$

$∴ D = B^{2} - 4 A C$

$\Rightarrow D = 81 (a + b)^{2} - 36 (2 a^{2} + 5 a b + 2 b^{2})$

$\Rightarrow D = 9 a^{2} + 9 b^{2} - 18 a b$

$\Rightarrow D = 9 (a - b)^{2} \geq 0$

$\Rightarrow D \geq 0$

So, the roots of the given equation are real and are given by:

$α = \frac{- B + \sqrt{D}}{2 A} = \frac{9 (a + b) + 3 (a - b)}{18} = \frac{12 a + 6 b}{18} = \frac{2 a + b}{3}$

and, $β = \frac{- B - \sqrt{D}}{2 A} = \frac{9 (a + b) - 3 (a - b)}{18} = \frac{6 a + 12 b}{18} = \frac{a + 2 b}{3}$

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