Q$)$Solve the equation : $x^{10}+11x^{5}+29=0$
Ans) My approach:
$x^{10}+11x^{5}+29=0$
Let's substitute $x^{5}=u$.
$\implies u^{2}+11u+29=0$
$\implies u=\frac{-11+\sqrt{5}}{2}$ and $u=\frac{-11-\sqrt{5}}{2}$
$\implies x^{5}=\frac{-11+\sqrt{5}}{2}$ and $x^{5}=\frac{-11-\sqrt{5}}{2}$
$\implies x^{5}-(\sqrt[5]{\frac{-11+\sqrt{5}}{2}})^{5}=0$ and $x^{5}-(\sqrt[5]{\frac{-11-\sqrt{5}}{2}})^{5}=0$
$\implies (x-\sqrt[5]{\frac{-11+\sqrt{5}}{2}})[x^{4}+x^{3}\sqrt[5]{\frac{-11+\sqrt{5}}{2}}+x^{2}(\sqrt[5]{\frac{-11+\sqrt{5}}{2}})^{2}+x(\sqrt[5]{\frac{-11+\sqrt{5}}{2}})^{3}+(\sqrt[5]{\frac{-11+\sqrt{5}}{2}})^{4}]=0$
and
$(x-\sqrt[5]{\frac{-11-\sqrt{5}}{2}})[x^{4}+x^{3}\sqrt[5]{\frac{-11-\sqrt{5}}{2}}+x^{2}(\sqrt[5]{\frac{-11-\sqrt{5}}{2}})^{2}+x(\sqrt[5]{\frac{-11-\sqrt{5}}{2}})^{3}+(\sqrt[5]{\frac{-11-\sqrt{5}}{2}})^{4}]=0$
Finally I can find only $2$ values of $x$ i.e. $x=\sqrt[5]{\frac{-11+\sqrt{5}}{2}}$ and $x=\sqrt[5]{\frac{-11-\sqrt{5}}{2}}$
I can't find out the remaining $8$ values of $x$. Please help me out. I am mainly facing problem in factorizing the expression to find the remaining $8$ values of $x$ .
The other $8$ values of $x$ are complex numbers and can be found via Euler's formula. If $r$ is any real number and $n$ an integer, then $r^{\frac{1}{n}}=|r^{\frac{1}{n}}|e^{i\frac{2k\pi}{n}}$ for integer values of $k$ such that $0\leq k\leq n-1$, which when collected constitute all the $n^{th}$ roots of $r$. You have found the roots corresponding to $k=0$ for both of your numbers. Now all you need to do is plug in the other values for $k$ into the formula. It may also help to know that $|r^{\frac{1}{n}}|e^{i\frac{2k\pi}{n}}=|r^{\frac{1}{n}}|(\operatorname{cos}(\frac{2k\pi}{n})+i\operatorname{sin}(\frac{2k\pi}{n}))$, to keep the results nice and tidy.