Question
$$\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x+4} = 0$$
Answer
$$ \begin{matrix}x_1 = -4-2 \sqrt{ 6 } & x_2 = -4+2 \sqrt{ 6 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1}{x}+\frac{1}{x-2}-\frac{1}{x+4} &= 0&& \text{multiply ALL terms by } \color{blue}{ x(x-2)(x+4) }. \\[1 em]x(x-2)(x+4)\cdot\frac{1}{x}+x(x-2)(x+4)\cdot\frac{1}{x-2}-x(x-2)(x+4)\cdot\frac{1}{x+4} &= x(x-2)(x+4)\cdot0&& \text{cancel out the denominators} \\[1 em]x^2+2x-8+x^2+4x-(x^2-2x) &= 0&& \text{simplify left side} \\[1 em]2x^2+6x-8-(x^2-2x) &= 0&& \\[1 em]2x^2+6x-8-x^2+2x &= 0&& \\[1 em]x^2+8x-8 &= 0&& \\[1 em] \end{aligned} $$
$ x^{2}+8x-8 = 0 $ is a quadratic equation.
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