I have this inequality
$$ (x-3)^2<5 \tag1 $$
But why is the following method wrong?
Add $-5$ to both sides and expand the square \begin{align} x^2-6x+9-5&<0 \\ x^2-6x+4&<0 \end{align} And completing the square of $x^2-6x+4<0$ gives: \begin{align} x^2-6x+ (6/2)^2-(6/2)^2+4&<0 \\ (x-3)^2-3^2+4&<0 \\ (x-3)^2&<5 \\ \end{align} Squaring and adding $3$ to both sides: \begin{align} x<3\pm\sqrt 5 \tag 2 \end{align} So I have: \begin{align} x&<3+\sqrt 5 \tag 3\\ x&<3-\sqrt 5 \tag 4 \end{align} Why is this method wrong?
From the book: For $(x-3)^2<5$ we have two solutions \begin{align} x-3&<\sqrt 5 \tag 5\\ -(x-3)&<\sqrt 5 \tag 6 \end{align} So \begin{align} x&<3+\sqrt 5 \tag 7\\ x&>3-\sqrt 5 \tag 8 \end{align}
$(7)$ is the same as my $(3)$, but $(4)$ is not $(8)$. What is wrong with my method?