$$\frac{5}{2x}+\frac{x+7}{x+4} = 2$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -\dfrac{ 9 }{ 5 }-\dfrac{\sqrt{ 91 }}{ 5 } & x_3 = -\dfrac{ 9 }{ 5 }+\dfrac{\sqrt{ 91 }}{ 5 } \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{5}{2x}+\frac{x+7}{x+4} &= 2&& \text{multiply ALL terms by } \color{blue}{ 2x(x+4) }. \\[1 em]2x(x+4)\cdot\frac{5}{2x}+2x(x+4)\frac{x+7}{x+4} &= 2x(x+4)\cdot2&& \text{cancel out the denominators} \\[1 em]5x^3+20x^2+2x^2+14x &= 4x^2+16x&& \text{simplify left side} \\[1 em]5x^3+22x^2+14x &= 4x^2+16x&& \text{move all terms to the left hand side } \\[1 em]5x^3+22x^2+14x-4x^2-16x &= 0&& \text{simplify left side} \\[1 em]5x^3+18x^2-2x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 5x^{3}+18x^{2}-2x = 0 } $, first we need to factor our $ x $.
$$ 5x^{3}+18x^{2}-2x = x \left( 5x^{2}+18x-2 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 5x^{2}+18x-2 = 0$.
$ 5x^{2}+18x-2 = 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.
This page was created using
Equations Solver