Find the distance of object from a concave mirror of focal length 10$cm$ so that image size is four times the size of the object.

When image is real:Real image is formed on the same side of the object, i.e. $u,v$ and $f$ all are negative.

So let

 $$\begin{gathered} u=-x \\ v=-4x as \left| \frac{v}{u}\right|=\left|m\right|=4 \\ and f=-10cm \end{gathered}$$

Substituting in $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

we have

 $$\begin{gathered} \frac{1}{-4x}-\frac{1}{x}=\frac{1}{-10} \\ \frac{5}{4x}=\frac{1}{10} \\ x=12.5cm \end{gathered}$$

When image is virtual:

In case of a mirror image is virtual, when it is formed behind the mirror, i.e. $u$ and $f$ are negative, while $v$ is positive.

So let

 $$\begin{gathered} u=-y \\ v=+4y \\ and f=-10cm \end{gathered}$$

Substituting in $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

we have

 $$\begin{gathered} \frac{1}{4y}-\frac{1}{y}=\frac{1}{-10} \\ \frac{3}{4y}=\frac{1}{10} \\ y=7.5cm \end{gathered}$$