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Emma Hancock Honors Algebra II 13 March 2014 Quadratic Equations: Assignment One 1. Find two consecutive whole numbers such that the sum of their squares is 265. 1st number: n 2nd number: n + 1 n2 + ( n + 1 )2 = 265 n2 + n2 + 2n + 1 = 265 2n2 + 2n - 264 = 0 2 ( n2 + n - 132 ) = 0 ( n2 + n - 132 ) / 2 = 0 / 2 n2 + n - 132 = 0 ( n - 11 ) ( n + 12) = 0 n - 11 = 0 n - 11 + 11 = 0 - 11 n = 11 n + 12 = 0 n + 12 - 12 = 0 - 12 n = -12 ( -12 is not a whole number so n must be positive 11) n + 1 = 11 + 11 n + 1 = 12 Solution: n = 11 and n + 11 = 12 Two consecutive whole numbers whose sum of squares is 265 are 11 and 12. 2. The perimeter of a rectangle is 44 inches and its area is 112 square inches. Find the length and the width of the rectangle. L = Length W = Width P = 2L + 2W 44 = 2L + 2W 44 - 2L = 2L - 2L + 2W 44 - 2L = 2W 44 - 2L / 2 = 2W / 2 22 - L = W W = 22 - L A=L*W 112 = L * W 112 = L * ( 22 - L ) 112 = 22L - L2 L2 - 22 L + 112 = 22L - 22L - L2 + L2
L2 - 22L + 112 = 0 ( L - 8 ) ( L - 14 ) = 0 L-8=0 L-8+8=0+8 L=8 L - 14 = 0 L - 14 + 14 = 0 + 14 L = 14 22 - 14 = W W=8 Solution: L = 14 inches and W = 8 inches A rectangle with a perimeter of 44 inches and an area of 112 has a length of 14 inches and a width of 8 inches. 3. The length of a rectangle is four meters more than twice its width. If the area of the rectangle is 126 square meters, find its length and width. L = Length W = Width L = 4 + 2W A=L*W 126 = W * ( 4 + 2W) 126 = 4W + 2W2 126 - 126 = 4W + 2W2 - 126 0 = 2W2 + 4W - 126 0 = 2 ( W2 + 2W - 63 ) 0 / 2 = (W2 + 2W - 63 ) / 2 0 = W2 + 2W - 63 0=(W-7)(W+9) W-7=0 W+7-7=0 +7 W=7 W+9=0 W+9-9=0-9 W = - 9 ( A dimension cannot be negative so W must be positive 7) L = 4 + 2W L = 4 + 2 (7) L = 4 + 14 L = 18 Solution: L = 18 meters and W = 7 meters The dimensions of a rectangle with an area of 126 meters and length four meters more than twice its width are 18 meters

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