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(male narrator) In Part 1 of this video,
we began looking at picture frames
and finding the width of a frame by drawing a picture.
Here, we have another problem
where an 8 by 12 inch drawing is put in a frame--
drawing a picture of an 8 by 12 inch drawing
put in a frame of uniform width.
We remember that the frame is on the left and right sides,
and so, we have an x
or an unknown distance on both sides.
Now the new width is 8 plus 2x.
Similarly, we have frame on top and bottom,
giving us 12 plus 2x.
We are told that the area of the frame
is equal to the area of the picture.
We'll have to do a little work to find out what this will be.
The picture itself is an 8 by 12 rectangle:
8 times 12 is equal to 96.
The picture has an area of 96...
which means that the frame, being the same,
must also have an area of 96.
All of it together in the large rectangle, then,
would be the picture, 96, plus the frame, 96,
giving us 192.
The total area, when we multiply the width by the length,
should equal that area of 192:
8 plus 2x, times 12, plus 2x, equals the 192.
We can start solving this by FOILing out what's left
to get 96, plus 16x, plus 24x, plus 4x squared, equals 192.
Combining like terms and putting things in order
gives us 4x squared, plus 40x, plus 96, equals 192.
In order to start solving, we want the equation to equal 0,
so we will subtract 192 from both sides.
This gives us 4x squared, plus 40x, minus 96, equals 0.
We are ready to start solving by factoring the equation.
Always start with the greatest common factor of 4,
leaving x squared, plus 10x, minus 24, equals 0.
We can continue factoring that trinomial
to be x, plus 12, times x, minus 2, equals 0.
To solve for x,
we take each factor with an x and set it equal to 0.
These equations solve quite nicely by subtracting 12
to tell us that x is equal to -12;
or adding 2 to tell us that x is equal to 2.
Recall that x is the width of the frame,
and so, we would not have a negative width on our frame.
Throwing the negative answer out,
the only possible width for our frame is a 2-inch frame.
As we set up our equations for our frame problems,
we must remember that the frame is on top and bottom,
so we have two x's;
and left and right, giving us two x's on that side as well.