Question
$$\frac{x+4}{x+5} = \frac{x-3}{2x}$$
Answer
$$ x = 0 $$
Explanation
$$ \begin{aligned} \frac{x+4}{x+5} &= \frac{x-3}{2x}&& \text{multiply ALL terms by } \color{blue}{ (x+5)\cdot2x }. \\[1 em](x+5)\cdot2x \cdot \frac{x+4}{x+5} &= (x+5)\cdot2x \cdot \frac{x-3}{2x}&& \text{cancel out the denominators} \\[1 em]2x^2+8x &= x^4+2x^3-15x^2&& \text{move all terms to the left hand side } \\[1 em]2x^2+8x-x^4-2x^3+15x^2 &= 0&& \text{simplify left side} \\[1 em]-x^4-2x^3+17x^2+8x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -x^{4}-2x^{3}+17x^{2}+8x = 0 } $, first we need to factor our $ x $.
$$ -x^{4}-2x^{3}+17x^{2}+8x = x \left( -x^{3}-2x^{2}+17x+8 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ -x^{3}-2x^{2}+17x+8 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using qubic formulas.
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