How to find all the solutions to $-8x^4-18x^2-11+\frac{1}{x^4}=0$? How to know all the roots for fourth degree  equation like this :
$$-8x^4-18x^2-11+\frac{1}{x^4}=0$$
 A: Just by the way, most people would not call this a degree 4 equation, because it also involves the reciprocal of $x$... only something like $4x^4-2x^3+x-5$ would be degree 4.
To answer your question:
First, note that the equation only involves even powers of $x$. Thus, if we let $t=x^2$, we can reduce the (apparent) complexity of the problem, and then once we've solved for $t$, the values of $x$ which are solutions will just be the various square roots of the values of $t$ which were solutions to the modified problem.
Letting $t=x^2$, we now want to solve for $t$ in
$$-8t^2-18t-11+\frac{1}{t^2}=0.$$
Now note that $t=0$ cannot possibly be a solution, because the left side involves a $\frac{1}{t}$. We have to be careful when multiplying both sides of an equation by a quantity involving the variable we're solving for, because sometimes it introduces fake solutions (as an illustration: if we had the equation $2=3$, which is false, multiplying both sides by $0$ gives $0=0$, which is true). Thus, even if $t=0$ appears to be a solution later, we know to throw it out.
Multiplying both sides by $t^2$ produces
$$-8t^4-18t^3-11t^2+1=0.$$
Now use the rational root theorem to check for possible solutions that are rational numbers. This won't work for every polynomial, but presumably, if these questions are being given to you in class, they will be designed so that the rational root theorem will help you.
In our problem, the rational root theorem tells us that, if there is some rational number $\frac{p}{q}$ (written in lowest terms, i.e. $p$ and $q$ have no common factors) such that $t=\frac{p}{q}$ is a solution, then we would have to have that
$$p\in\{\pm 1\},\quad q\in\{\pm 1,\pm 2,\pm 4,\pm 8\}.$$
Checking each of the numbers
$$\pm 1,\quad \pm \frac{1}{2},\quad \pm \frac{1}{4},\quad\pm \frac{1}{8}$$
to see whether they are solutions, we see that there are 3 rational number solutions to
$$-8t^4-18t^3-11t^2+1=0,$$
namely $t=-1$, $t=-\frac{1}{2}$, and $t=\frac{1}{4}$. Now, we know that a degree $n$ polynomial can have at most $n$ distinct roots; here, we have a degree 4 polynomial, and we found 3 roots that were rational numbers. Perhaps there is another root that is an irrational number? In fact, this is not the case here. The root $t=-1$ is a double root. You can check this by using whatever polynomial division algorithm you've learned to successively divide the polynomial $-8t^4-18t^3-11t^2+1$ by $t-(-1)$, $t-(-\frac{1}{2})$, and $t-(\frac{1}{4})$, and then noting that $-1$ is a root of the degree 1 polynomial that's left over. 
Thus, we've determined that the only solutions to
$$-8t^4-18t^3-11t^2+1=0$$
are $t=-1$, $t=-\frac{1}{2}$, and $t=\frac{1}{4}$. Now you need to go back to the original question, namely, finding the solutions to
$$-8x^8-18x^6-11x^4+1=0.$$
Remember, since we set $t=x^2$, if $t$ is a solution to the equation involving $t$, then the two different square roots of $t$, namely $\sqrt{t}$ and $-\sqrt{t}$, will both be solutions for $x$ in the equation involving $x$.
Therefore, the solutions to
$$-8x^8-18x^6-11x^4+1=0$$
are
$$\begin{align*}
x&=\sqrt{-1}=i, &  x&=\sqrt{-\frac{1}{2}}=\frac{i\sqrt{2}}{2},& x=\sqrt{\frac{1}{4}}=\frac{1}{2},\\\\\\
x&=-\sqrt{-1}=-i, & x&=-\sqrt{-\frac{1}{2}}=-\frac{i\sqrt{2}}{2}, & x=-\sqrt{\frac{1}{4}}=-\frac{1}{2}
\end{align*}$$
You should note that because $t=-1$ was a double root of $-8t^4-18t^3-11t^2+1=0$, the solutions for $x$ that arose from it, namely $x=i$ and $x=-i$, will each be double roots of $-8x^8-18x^6-11x^4+1=0$. Thus, we have found a total of 8 (counting with muliplicity) solutions to a degree 8 equation, and therefore we must have got all of them. We're done :)
A: Rewrite as $8x^8+18x^6+11x^4-1=0$. By inspection, $i$ is a root, and therefore so is $-i$. Divide the polynomial by $x^2+1$. We get $8x^6+10x^4+x^2-1$. Note that $i$ is a root of this, and therefore so is $-i$. Divide again by $x^2+1$. We get $8x^4+2x^2-1$. Let $y=x^2$. Now we have an easy quadratic.
Remark: Please note the comment by lab bhattacharjee. Things are easier still if we let $x^2=y$ immediately. Then we are looking at $8y^4+18y^3+11y^2-1=0$, and the root $-1$ is obvious.  Divide by $y+1$, and continue. 
A: You can multiply both sides by $x^4$ to get a $4^\text{th}$ degree equation in $x^2$. There are formulas using addition, subtraction, radicals etc for all polynomials up to $4^{\text{th}}$ degree. It is proved that no such formula exists for higher degree polynomials. You can look here for info on the quartic formula if interested http://www.sosmath.com/algebra/factor/fac12/fac12.html. But other practical ways to find roots are simply by trying different values of $x$ or you can also plug in values of $x^2$ in this case. Once you get a root $r$ you can divide the polynomial by $x - r$ to be left with a simpler polynomial to find the rest of the roots. In your problem though, I don't see any straight forward roots.
André Nicolas showed that x = i works, so you can proceed from there.
