Question
$$2 \cdot \frac{x}{x^3-5x}+\frac{2}{x^2+5x} = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -\dfrac{ 5 }{ 4 }-\dfrac{\sqrt{ 65 }}{ 4 } & x_3 = -\dfrac{ 5 }{ 4 }+\dfrac{\sqrt{ 65 }}{ 4 } \end{matrix} $$
Explanation
$$ \begin{aligned} 2 \cdot \frac{x}{x^3-5x}+\frac{2}{x^2+5x} &= 0&& \text{multiply ALL terms by } \color{blue}{ (x^3-5x)(x^2+5x) }. \\[1 em](x^3-5x)(x^2+5x)\cdot2 \cdot \frac{x}{x^3-5x}+(x^3-5x)(x^2+5x)\cdot\frac{2}{x^2+5x} &= (x^3-5x)(x^2+5x)\cdot0&& \text{cancel out the denominators} \\[1 em]2x^3+10x^2+2x^3-10x &= 0&& \text{simplify left side} \\[1 em]4x^3+10x^2-10x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 4x^{3}+10x^{2}-10x = 0 } $, first we need to factor our $ x $.
$$ 4x^{3}+10x^{2}-10x = x \left( 4x^{2}+10x-10 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 4x^{2}+10x-10 = 0$.
$ 4x^{2}+10x-10 = 0 $ is a quadratic equation.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.
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