$$\frac{1}{4}x^4-\frac{1}{2}x^2 = x\cdot2$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 2.33075 & x_3 = -1.16537+1.44024i \\[1 em] x_4 = -1.16537-1.44024i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1}{4}x^4-\frac{1}{2}x^2 &= x\cdot2&& \text{multiply ALL terms by } \color{blue}{ 4 }. \\[1 em]4 \cdot \frac{1}{4}x^4-4\frac{1}{2}x^2 &= 4x\cdot2&& \text{cancel out the denominators} \\[1 em]x^4-2x^2 &= 8x&& \text{move all terms to the left hand side } \\[1 em]x^4-2x^2-8x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{4}-2x^{2}-8x = 0 } $, first we need to factor our $ x $.
$$ x^{4}-2x^{2}-8x = x \left( x^{3}-2x-8 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ x^{3}-2x-8 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using qubic formulas.
This page was created using
Polynomial Equations Solver