solve following quadratic equation : 2x/x-4 + 1/x-2 + 2/x2-6x+8 =0

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Solution

$\frac{2 x}{x - 4} + \frac{1}{x - 2} + \frac{2}{x^{2} - 6 x + 8} = 0 \Rightarrow \frac{2 x}{x - 4} + \frac{1}{x - 2} = - \frac{2}{x^{2} - 6 x + 8} \Rightarrow \frac{2 x (x - 2) + (x - 4)}{(x - 4) (x - 2)} = - \frac{2}{x^{2} - 6 x + 8} \Rightarrow \frac{2 x^{2} - 4 x + x - 4}{x^{2} - 6 x + 8} = - \frac{2}{x^{2} - 6 x + 8} \Rightarrow 2 x^{2} - 3 x - 4 = - 2 a n d x^{2} - 6 x + 8 \neq 0 \Rightarrow 2 x^{2} - 3 x - 2 = 0 a n d x^{2} - 6 x + 8 \neq 0 \Rightarrow 2 x (x - 2) + 1 (x - 2) = 0 a n d x (x - 2) - 4 (x - 2) \neq 0 \Rightarrow (x - 2) (2 x + 1) = 0 a n d (x - 2) (x - 4) \neq 0 \Rightarrow x = 2, x = \frac{- 1}{2} a n d x \neq 2, x \neq 4 ∴ x = - \frac{1}{2}$

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