7/9/2013 Quadratic Equations 43 Expanding a Room Two methods: A = A 1 + A 2 + A 3 + A 4 A = (9 + x)(12 + x) = 216 OR Either way: x 2 + 21x + 108 = 216 Sox 2 + 21x – 108 = 0 ≈ 4.3 sq.
{"Image DocumentName": "Solving Quadratic Equations", "Image ImageDescription": "Expanding a Room. Solving Quadratic Equations. Two methods: A = (9 + x)(12 + x) = 216. OR. A = A1 + A2 + A3 + A4. = 216. Either way: x2 + 21x = 216. So. x2 + 21x \u2013 108 = 0. x. = 2 1. ( ) 21 2 \u2013 4 1 \u2013108. \uf0d6 21. \u2013 x. = \uf0d6 21. \u2013 Expanding A Room. Here we apply our knowledge of the quadratic formula to the solution of a practical problem \u2013 how much to expand a room by to meet certain dimension conditions. We wish to double the area A1 of the room by expanding it x ft in both length and width. The new area A is thus 2A1. Since A1 = 108 then A = 2A1 = 216. As the illustration indicates there are two ways to do this: by adding up the original area, together with the three new areas formed by the new rectangles, that is, A = A1 + A2 + A3 + A4. Or we can consider the new outside dimensions of the room 9 + x and 12 + x, and simply find the product of the two: A = (9 + x)(12 + x) which yields the same final equation to solve: x2 + 21x \u2013 108 = 0. It doesn\u2019t take long to figure out that this polynomial is not factorable, so we resort to using the quadratic formula. We discard the negative root as being not a real measure of length and so choose the positive root, approximately 4.3 feet. \u2248 4.3 sq. ft. 7/9/2013. Quadratic Equations. Quadratic Equations. 7/9/2013.", "Image XResolution": "37", "Image YResolution": "37", "Image ResolutionUnit": "Pixels/Centimeter", "Image Software": "SlidePlayer", "Image YCbCrPositioning": "Centered"}