A rectangular photograph is twice as tall as it is wide. 
*

*A rectangular photograph is twice as tall as it is wide. When a 4 cm wide frame is placed 
around the photo, the area of the frame and photo equals three times that of the photo alone. 
The original width of the photograph alone is: 
a) 4 cm 
b) 6 cm 
c) 8 cm 
d) 10 cm 
e) 12 cm
 A: You have a photograph that measures $w$ by $2w$, thus the area of the photograph is $2w^2$. The frame is 4cm wide which adds 4 cm on the left, top, bottom, and right. So the total frame measures $w+8$ by $2w+8$, with an area of $(w+8)(2w+8)$. We know the following equality:
$$(w+8)(2w+8)=3(2w^2)$$
$$2w^2+24w+64=6w^2$$
$$4w^2-24w-64=0$$
$$w^2-6w-16=0$$
$$(w-8)(w+2)=0$$
And the only positive solution is $w=8$ so the answer is c.
The answer is very similar to Martin's answer, however, he forgets to include the 4cm on both sides of the frame:
 _________
|  _____  |
| |     | |
| |_____| |
|____w____|
4cm     4cm

A: Let the photograph be $x$ cm wide and $2x$ cm tall.
You have
\begin{equation} (x+4)(2x+4) = (2x^2)\cdot3, \end{equation}
as the first side has now length $x+4$ and the other one has length $2x+4$. The area before placing the frame was $2x^2$.
Now you just have to solve the equation:
\begin{equation} (x+4)(2x+4) = (2x^2)\cdot3, \end{equation}
\begin{equation} 2x^2 + 12x + 16 = 6x^2, \end{equation}
\begin{equation} 4x^2 - 12x - 16 = 0, \end{equation}
\begin{equation} x^2 - 3x - 4 = 0, \end{equation}
\begin{equation} (x-4)(x+1) = 0, \end{equation}
giving you answer a) 4cm.
