Inverses of elements for (Part One) an ideal generated by a reducible polynomial and (Part Two) an irreducible polynomial with no real roots

Question

Let $\langle x^2-4 \rangle$ denote the ideal generated by the polynomial $x^2-4$. (Artin uses $(x^2-4)$.)

In connection with this Visualizing quotient polynomial rings are fields for maximal ideals which are generated by irreducible monic

Part One: Inverses in the quotient ring $\mathbb R[x]/\langle x^2-4 \rangle$.

1. Consider the quotient ring $\mathbb R[x]/\langle x^2-4 \rangle$ and one of its elements $\overline {x^2-2x} = [x^2+2x+\langle x^2-4 \rangle]$. I think this element has no multiplicative inverse because $x^2-2x=0$ for $x=2$. Is that correct?

2. In the quotient ring, $\mathbb R[x]/\langle x-2 \rangle$, the relation is "x=2". This relation is used to compute inverses. What exactly is the relation in the quotient ring $\mathbb R[x]/\langle x^2-4 \rangle$? I think it is "$x=2$ or $x=-2$".

3. In the quotient ring $\mathbb R[x]/\langle x^2-4 \rangle$, does the element $\overline {x+1} = [x+1+\langle x^2-4 \rangle]$ have an inverse, and why? If so, then what is it? (I think I can answer these if I know the answer for 2.)

Part Two: Inverses in the quotient ring $\mathbb R[x]/\langle x^2+1 \rangle$.

For the quotient ring $\mathbb R[x]/\langle x-2 \rangle$, the relation is $x=2$ so inverse of $\overline p = [p(x)+\langle x-2 \rangle]$ is $\overline {\frac{1}{p(2)}}$.

For the quotient ring $\mathbb R[x]/\langle x^2+1 \rangle$, I believe $x^2+1$ is irreducible in $\mathbb R[x]$, so $\mathbb R[x]/\langle x^2+1 \rangle$ is a field.

To compute the inverse of $\overline p = [p(x)+\langle x^2+1 \rangle]$, I believe I need the relation in the quotient ring $\mathbb R[x]/\langle x^2+1 \rangle$.

1. What's the relation in the quotient ring $\mathbb R[x]/\langle x^2+1 \rangle$? I think it is "$x= i$ or $x=-i$".

2. What is the inverse of $\overline p = [p(x)+\langle x^2+1 \rangle]$?

( I think I will know the answer if I know the answer to 2 in Part One.)