Question about the properties of radicals all. I was solving a simple problem when I realized I had something seemingly contradictory come up. I completed the square on a polynomial as such:
$$
\begin{eqnarray}
y^2-3y&=&-1\\
y^2-3y+\frac{9}{4}&=&-1+\frac{9}{4}\\
(y-\frac{3}{2})^2&=&\frac{5}{4}\\
y-\frac{3}{2}&=& \pm \sqrt{\frac{5}{4}}\\
y&=& \frac{3}{2} \pm \sqrt{\frac{5}{4}}\\
y&=& \frac{3}{2} \pm \frac{\sqrt{5}}{2}\\
y&=& \frac{3 \pm \sqrt{5}}{2}
\end{eqnarray}
$$
However, here it seems that I got confused. Knowing the Square Root Property, the solution set of $x^2 = k \hspace{1mm}is \hspace{1mm}\{\pm \sqrt{k}\}.$
But in other instances, the radical sign is used to designate the principal square root, which only is used to designate a nonnegative square root.
Which leads to my main question. Ignoring the positive root of 4, I decided to take the negative square root of 4 as such:
$$
\begin{eqnarray}
y-\frac{3}{2}&=& \pm \sqrt{\frac{5}{4}}\\
y&\stackrel{?}{=}& \frac{3}{2} \pm \frac{\sqrt{5}}{-2} \hspace{2 cm}(\sqrt{4}=-2)\\
y&\stackrel{?}{=}& \frac{3 \pm \sqrt{5}}{-2}
\end{eqnarray}
$$
Plugging $y=\frac{3 \pm \sqrt{5}}{-2}$ back into $y^2-3y$, yields $8+3\sqrt{5}, 8-3\sqrt{5}$, respectively, so,
$$
\begin{eqnarray}
y&\neq& \frac{3 \pm \sqrt{5}}{-2}
\end{eqnarray}
$$
I'm wondering where my reasoning fell short and how I can fix it when I come across something similar in the future. Why are we assuming the square root of 4 in the denominator to be a positive square root, i.e., the principal square root, and why does the equation become false when we take the negative square root? Thank you, I greatly appreciate it.

The way you are adding fractions is not correct. Note that$$\frac32+\frac{\sqrt5}{-2}=\frac32+\frac{-\sqrt5}2=\frac{3-\sqrt5}2$$and that$$\frac32-\frac{\sqrt5}{-2}=\frac32+\frac{\sqrt5}2=\frac{3+\sqrt5}2.$$So, you get$$\frac32\pm\frac{\sqrt5}{-2}=\frac{3\mp\sqrt5}2,$$which are the same numbers that you had got before.
