Frames – Picture Frames (part 2) – YouTube.mp4

(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. 