The sign tells you which side

The image in a flat mirror is as far behind the mirror as the object stands in front of it. It is not on the glass, the way a picture sits on a wall.

Confirm it by tracing two reflected rays backward to where they appear to meet: that point lies behind the mirror. Then read off the consequences the on-the-glass picture misses:

1. Stand $2$ m from the mirror and your image is $2$ m behind it, so $4$ m from your eyes. That is why you have to focus as though looking at something across the room.
2. Step back to $3$ m and the image retreats to $3$ m behind, now $6$ m away. Object-to-image separation is always twice your distance to the mirror.
3. A mirror half your height, correctly positioned, shows all of you at any distance, which is a geometry result the on-the-glass picture cannot produce.

$$\frac{1}{s_i} + \frac{1}{s_o} = \frac{1}{f}.$$

Solve it and $s_i$ can come out negative. That is the answer, not a failure: a negative image distance puts the image on the side the light never actually reaches, which is what virtual means.

So $s_i = -12$ cm means the image is $12$ cm behind the mirror, virtual and upright. The magnitude gives the distance; the sign gives the side. Stripping the sign and reporting a real image $12$ cm in front of the mirror describes something no screen could ever catch.

One more piece of algebraic discipline goes with it. The equation is a sum of reciprocals, so stay in reciprocal space until the very end: reach $1/s_i$, then invert once. Reporting $1/s_i$ as the distance produces an image a fraction of a centimetre away, which is the tell.

The sign of $f$ is decided by the optic, not by the number printed in the problem.

1. Concave mirror: converging. $f > 0$, and $f = R/2$.
2. Convex mirror: diverging. $f < 0$, because its focal point sits on the side the light never reaches.

So a convex mirror described as having "focal length $20$ cm" enters the equation as $f = -20$ cm. Use $+20$ and you get a real image from a convex mirror, which cannot happen.

That impossibility is the check worth keeping. A diverging mirror gives an upright, reduced, virtual image for every object distance, with no exceptions. If your arithmetic produces anything else from a convex mirror, the sign of $f$ went in wrong. That fixed outcome is also why convex mirrors are used for wide-angle security and passenger-side car mirrors: reduced means more of the scene fits.

The CED writes the magnification with bars on every term:

$$|M| = \left|\frac{h_i}{h_o}\right| = \left|\frac{s_i}{s_o}\right|,$$

which gives the size ratio: above one is enlarged, below one is reduced. Orientation is a separate reading, and this course carries it in the sign of $M = h_i/h_o$: negative means inverted, positive means upright.

So $M = -2$ splits into two statements: the image is twice as tall as the object, and it is inverted. It says nothing at all about which side of the mirror the image is on: that is the job of $s_i$.

Two further checks. $M = 0.5$ means reduced and upright, not inverted, since the sign and the magnitude answer different questions. And for a single mirror or lens, real goes with inverted and virtual goes with upright, so $M$ and $s_i$ should always agree with each other. When they disagree, one of them was computed wrongly.