Somehow I'm supposed to solve this by creating a quadratic equation and factoring it. The question is: A photograph measuring 12 by 8cm is to be surrounded by a mat before framing. The width of the mat is to be the same on all sides of the photograph. The area of the mat is to equal the area of the photograph. Find the width of the mat. Please explain if you answer, I really dont get this. YOU WANT REP? I HAS REP TO GIVE YOU.
Lol, you could've just searched the problem on google xD Seems like it's all over it @_@ Good luck =) Edit: saw your question on yahoo :B You got a good answer there already, but google showed a few more: http://www.physicsforums.com/showthread.php?t=24333 http://answers.yahoo.com/question/index ... 209AArP2nE
What you probably mean is that the area of the mat not covered by the picture (the area of the frame) is equal to the area of the picture length of the picture = L = 12 height of the picture = H = 8 area of the picture = L*H = 12*8 = 96 width of the frame = W = ? length of the frame = 2*W + L height of the frame = 2*W + H area of the frame = (2*W + L)*(2*W + H) - 96 = (2*W + 12)*(2*W + 8) -96 = (4*W^2 + 40*W + 96) - 96 = 4*W^2 + 40*W set area of frame = area of the picture 4*W^2 + 40*W = 96 4*W^2 + 40*W - 96 = 0 W = 2cm EDIT: here is the factorization of the quadratic equation: 4*W^2 + 40*W - 96 = 0 4*(W^2 + 10*W - 24) = 0 (W^2 + 10*W - 24) = 0 (W+12)*(W-2) = 0 W = -12 or 2 W cannot be negative, so W = 2
Do you know what the answer is? I got a really absurd number (I haven't done quads since middle school) :x I keep getting 2cm
As you see from my post, 2cm is the correct answer. This is not absurd. This means that the mat has dimensions 14x10 and the picture is 12x8...
Okay I'm sorry haha I though it was 2meters lollll I don't use metric units i'm from the US, just a careless mistake. ^^used the same reasoning as you guys did smartypants in here Edit: the USA really needs to switch, it's frustrating...that's the reason one of the space shuttles was wacky some scientist took cm to be inches -_-
You know, you can just input the numbers into a graphing calculator and find the intersections. (It works for me...i have the new T-I Nspire)