$$\frac{1}{x^2+2x}+\frac{x-1}{x} = 1$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -1 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1}{x^2+2x}+\frac{x-1}{x} &= 1&& \text{multiply ALL terms by } \color{blue}{ (x^2+2x)x }. \\[1 em](x^2+2x)x\cdot\frac{1}{x^2+2x}+(x^2+2x)x\frac{x-1}{x} &= (x^2+2x)x\cdot1&& \text{cancel out the denominators} \\[1 em]x+x^3+x^2-2x &= x^3+2x^2&& \text{simplify left side} \\[1 em]x^3+x^2-x &= x^3+2x^2&& \text{move all terms to the left hand side } \\[1 em]x^3+x^2-x-x^3-2x^2 &= 0&& \text{simplify left side} \\[1 em]x^3+x^2-x-x^3-2x^2 &= 0&& \\[1 em]-x^2-x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -x^{2}-x = 0 } $, first we need to factor our $ x $.
$$ -x^{2}-x = x \left( -x-1 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ -x-1 = 0$.
This page was created using
Polynomial Equations Solver