Answer
They chose different signs for the magnification.
Work Step by Step
Whether the object was to be upright or inverted was not specified, and one student chose m = 3 while the other chose m = -3, that is, one assumed the image was upright, and the other student assumed the image was inverted.
Relate the object and image distances to the magnification.
$$m=-\frac{d_i}{d_o}$$
If m is positive
$$d_i=-3d_o$$
Use the mirror equation, 23–2. The focal length of the lens is f = 12 cm.
$$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$
Solve for the object distance.
$$\frac{1}{d_o}+\frac{1}{-3d_o}=\frac{1}{12cm}$$
$$d_o=8.0cm$$
If m is negative
$$d_i=3d_o$$
Use the mirror equation, 23–2. The focal length of the lens is f = 12 cm.
$$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$$
Solve for the object distance.
$$\frac{1}{d_o}+\frac{1}{3d_o}=\frac{1}{12cm}$$
$$d_o=16cm$$
The images produced are virtual and real, respectively.