Question
$$16x^5-20x^3+5x+\frac{6^{1/2}}{4}-\frac{2^{1/2}}{4} = 0$$
Answer
$$ x = 0 $$
Explanation
$$ \begin{aligned} 16x^5-20x^3+5x+\frac{6^{1/2}}{4}-\frac{2^{1/2}}{4} &= 0&& \text{multiply ALL terms by } \color{blue}{ 4 }. \\[1 em]4\cdot16x^5-4\cdot20x^3+4\cdot5x+4(\frac{6^{1/2}}{4}-\frac{2^{1/2}}{4}) &= 4\cdot0&& \text{cancel out the denominators} \\[1 em]64x^5-80x^3+20x+0 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 64x^{5}-80x^{3}+20x = 0 } $, first we need to factor our $ x $.
$$ 64x^{5}-80x^{3}+20x = x \left( 64x^{4}-80x^{2}+20 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 64x^{4}-80x^{2}+20 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using quartic formulas
This page was created using
Polynomial Equations Solver