= Solution
Use positive object distance $u=20\,\mathrm{cm}$ and <focal length> $f=4\,\mathrm{cm}$. The <thin lens> equation gives
$$
\frac1v=\frac14-\frac1{20}=\frac15\,\mathrm{cm^{-1}},\qquad\boxed{v=5.0\,\mathrm{cm}.}
$$
The <transverse magnification> is $M=-v/u=-1/4$, so the signed height is
$$
\boxed{h_i=Mh_o=-\frac14(2\,\mathrm{cm})=-0.50\,\mathrm{cm}.}
$$
The <real image> lies \b[$5.0\,\mathrm{cm}$ beyond the lens], is \b[inverted], and has positive height magnitude \b[$0.50\,\mathrm{cm}$].
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