Roots of Polynomial Equation? 
$y=1/x$ so I plugged in $x=1/y$ into the equation above and got $y^{4}+y^{3}+y^{2}/c+y/4-1/2$, but apparently it's wrong, when I looked up the answer below.

What am I missing? 
 A: It looks like you didn't substitute $1/y$ in for $x$; rather you seem to have simply replaced every instance of an $x$ with a $y$ and then replaced the numerical coefficients with their reciprocals.
If you do the substitution $x=1/y$ you end up with:
$$(1/y)^4 + (1/y)^3  + c(1/y)^2 + 4(1/y) - 2 = 0$$
Now, it's possible that you got this far correctly, and then incorrectly took the reciprocal term-by-term.  What you ought to do is multiply through by $y^4$ to clear the denominators out.
A: The other two answers to your question already tell you what you did wrong, so I'll just leave a note about why this subsitution gives the reciprocal roots.

If a polynomial $P : \mathbb{R}\rightarrow\mathbb{R}$ with $P(x)=0$ has exactly four roots $\alpha,\, \beta,\, \gamma,\,$and $\delta,\,$there must exist an $\epsilon \in \mathbb{R}$ such that $$P(x)=\epsilon \, (x-\alpha)(x-\beta)(x-\gamma)(x-\delta)$$
To find a rational function $Q(y)$ with roots $1/\alpha$, $1/\beta$, $1/\gamma$, and $1/\delta$, we can use the substitution that you're supposed to use, and set $Q=0$:
$$Q(y)=\epsilon \, \left(\frac{1}{y}-\alpha\right)\left(\frac{1}{y}-\beta\right)\left(\frac{1}{y}-\gamma\right)\left(\frac{1}{y}-\delta\right)=0$$
You might already be able to see that this gives the reciprocal roots. To make this a polynomial, we can multiply by $y^4$ on both sides, distributing one $y$ to each of the factors:
$$y^4 \cdot Q(y) = \epsilon \, \left(1-y\alpha\right)\left(1-y\beta\right)\left(1-y\gamma\right)\left(1-y\delta\right)=0$$
It's now clear that the roots of the polynomial $y^4 \cdot Q(y)$ are the reciprocals of the roots of $P(x)$. 

Even more simply, if $P(x)=0$ when $x=\alpha$, then $P(1/x)=0$ when $x=1/\alpha$. The only additional step is multiplying by powers of the variable to make a polynomial from $P(1/x)$. 
A: $$\begin{align}
x^4 + x^3 + cx^2 + 4x - 2 & = 0 \\
\left(\frac{1}{y}\right)^4 + \left(\frac{1}{y}\right)^3 + c\left(\frac{1}{y}\right)^2 + 4\left(\frac{1}{y}\right) - 2 & = 0 \\
\frac{1}{y^4} + \frac{1}{y^3} + c\frac{1}{y^2} + 4\frac{1}{y} - 2 & = 0 \\
\text{multiply by }y^4\text{ since y=0 isn't a solution} \\
1 + y + cy^2 + 4y^3 - 2y^4 & = 0 \\
\text{multiply by -1 and rearrange to get the same answer as the book}
\end{align}$$
