Page 204

messersmith_power_intermediate_algebra_1e_ch4_7_10

Step 4: Solve the equation. 252 (x 1)(x 4)(2) 126 (x 1)(x 4) Divide both sides by 2. 126 x2 3x 4 Multiply. 0 x2 3x 130 Write in standard form. 0 (x 10)(x 13) Factor. x 10 0 or x 13 0 Set each factor equal to zero. x 10 or x 13 Solve. Step 5: Check the answer, and interpret the solution as it relates to the problem. Because x represents the width, it cannot be negative. Therefore, the width of the original piece of cardboard is 13 in. The length of the cardboard is x 5, so 13 5 18 in. Width of cardboard 13 in.    Length of cardboard 18 in. Check: Width of box 13 4 9 in.; Length of box 13 1 14 in.; Height of box 2 in. Volume of box 9(14)(2) 252 in3. YOU TRY 3 The width of a rectangular piece of cardboard is 2 in. less than its length. A square piece that measures 3 in. on each side is cut from each corner, then the sides are turned up to make a box with volume 504 in3. Find the length and width of the original piece of cardboard. 3 Solve an Applied Problem Involving Area EXAMPLE 4 A rectangular pond is 20 ft long and 12 ft wide. The pond is bordered by a strip of grass of uniform (the same) width. The area of the grass is 320 ft2. How wide is the border of grass around the pond? Solution Step 1: Read the problem carefully. Draw a picture. Step 2: Choose a variable to represent the unknown, and defi ne the other unknowns in terms of this variable. 2 2x x 12 2 x x x x width of the strip of grass 20 2x length of pond plus two strips of grass 12 2x width of pond plus two strips of grass 12 2x 648 CHAPTER 10 Quadratic Equations and Functions www.mhhe.com/messersmith


messersmith_power_intermediate_algebra_1e_ch4_7_10
To see the actual publication please follow the link above