So to solve this type of quadratic, the x–coefficient wouldn't be the perimeter of a rectangle but would be related to the difference in the lengths. Then with ( x - α ) ( x + β ) = x 2 - b x - c we have that α - β = b and α β = c. To get a quadratic of the form x 2 - b x - c we need the roots to be of opposite signs. However, one of the form x 2 - b x - c is a little more complicated. This simply means that the ruler was the wrong way around. If we start instead with x 2 + b x + c then this is like having a rectangle with both width and height negative. The above strategy can be used to solve a quadratic of the form: x 2 - b x + c = 0 with b and c positive. The final dimensions of this rectangle are: This has perimeter P and area A and can be rearranged into a rectangle without changing its perimeter or area. So we remove a square of side length P 2 16 - A to make a (symmetric) L–shape. This square has area P 2 16, which is P 2 16 - A too much. To get a square with perimeter P we need side length P 4. We can run this algorithm from an arbitrary starting point, i.e., perimeter P and area A. The obvious thing to do is to draw the squares. We want to put those together in a way that makes the story clear. Geometrically, this again means finding a square but this time with area P 2 16 - A.Ī square with the same perimeter as the original rectangle,Ī second square whose area is the difference between the new square and the original rectangle. If we then tried to solve for w, we would square root P 2 16 - A. Looking at Equation ( 1), the quantity P 2 16 is then the area of that square and P 2 16 - A is the difference between the area of the square and the original rectangle. So P 4 is the side length of a square with the same perimeter as our original rectangle. If it were the perimeter of a square then P 4 would be the side length. This is quite straightforward to come up with since the quantity P is the perimeter of a rectangle. So to find a route to Equation ( 1), we need to find a justification for the quantity P 4. Key to Equation ( 1) is the quantity P 4. The goal is to find a way to effectively get that formula without doing the algebra or mentioning the phrase "quadratic equation". So if we know A and P, we can get an equation for w by eliminating h: The equations linking area, A, perimeter, P, height, h, and width, w, are: The quadratic equation for the width is simple to set up. The level at which I was aiming this activity meant that I didn't want to make that explicit, but wondered if I could figure out how to make solving it a natural progression. Obviously, starting with the width and height then one can easily calculate the perimeter and width, but going the other way involves solving a quadratic. My aim was to create an activity with the various formulae relating the width and height of a rectangle to its perimeter and area. I was playing around with formulae involving rectangles and figured out a different take on the geometry of completing the square.
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