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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