Help...I'm stuck: "Marcus is unhappy with his 2-inch-wide frame...."

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crreedjr

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Hello- I need help with the problem below.....

"Marcus is unhappy with his 2-inch-wide frame, so he decides to put a frame with a different width around his canvas. The total area of the canvas and the new frame is given by the polynomial w^2 - w - 2 where w represents the width of the canvas. Determine the width of the new frame. Show all your work. Explain WHY you did each step."

Here is what I have so far..."First, you factor w^2-w-2 to simplify it. Now, the equation is (w-2)(w+1)", but now I'm stuck. Thanks!
 
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crreedjr said:
"Marcus is unhappy with his 2-inch-wide frame, so he decides to put a frame with a different width around his canvas. The total area of the canvas and the new frame is given by the polynomial w^2 - w - 2 where w represents the width of the canvas. Determine the width of the new frame. Show all your work. Explain WHY you did each step."

Here is what I have so far..."First, you factor w^2-w-2 to simplify it. Now, the equation is (w-2)(w+1)", but now I'm stuck.
Click to expand...
As currently stated, I see no reason for the mention of the previous frame (since that's been removed from consideration and then replaced with a different frame), and I see no way to solve this, because there is no enough information.

Is this maybe a follow-up to a previous exercise? Is there maybe a picture that goes with this exercise, maybe something that gives more information about the canvas? Thank you! ;)
 
Initial problem statement.....

stapel said:
As currently stated, I see no reason for the mention of the previous frame (since that's been removed from consideration and then replaced with a different frame), and I see no way to solve this, because there is no enough information.

Is this maybe a follow-up to a previous exercise? Is there maybe a picture that goes with this exercise, maybe something that gives more information about the canvas? Thank you! ;)
Click to expand...

area%20directions.png
 
crreedjr said:
area%20directions.png

"Marcus is unhappy with his 2-inch-wide frame, so he decides to put a frame with a different width around his canvas. The total area of the canvas and the new frame is given by the polynomial w^2 - w - 2 where w represents the width of the canvas. Determine the width of the new frame. Show all your work. Explain WHY you did each step."

Here is what I have so far..."First, you factor w^2-w-2 to simplify it. Now, the equation is (w-2)(w+1)", but now I'm stuck. Thanks!
Click to expand...
Click to expand...

The area before framing = (w-3) * w

The area after framing = (w-2) * (w+1)

Does that give a clue regarding the width of the new frame?

How did you arrive at that?

Write it up.......
 
Subhotosh Khan said:
The area before framing = (w-3) * w

The area after framing = (w-2) * (w+1)

Does that give a clue regarding the width of the new frame?

How did you arrive at that?

Write it up.......
Click to expand...


Ok, so....area before framing in (w-3)w or w^2-3w and area after framing is w^2-w-2, if I subtract w^2-w-2 from w^2-3w I get -2w+2...solve for w....w=1

So the width of the new frame is 1 inch, correct?
 
crreedjr said:
Ok, so....area before framing in (w-3)w or w^2-3w and area after framing is w^2-w-2, if I subtract w^2-w-2 from w^2-3w I get -2w+2...solve for w....w=1

So the width of the new frame is 1 inch, correct?
Click to expand...
No...

suppose the width of the frame = x inch

then

area after framing = (w + 2x)(w - 3 + 2x).......why?

so

(w + 2x)(w - 3 + 2x) = w2 - w - 2

Now solve for 'x'
 
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