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3 Objectives 1.Solve an integer application using a quadratic equation. 2.Solve a motion application using a quadratic equation. 3.Solve a geometric application using a quadratic equation. 1 1 2 2 3 3

4 Solve an integer application using a quadratic equation 1.

5 Example One integer is 5 less than another and their product is 84. Find the integers. 1.What am I asked to find? The larger number: x The smaller number: x - 5 2.Form an equation x(x – 5) = 84

6 Example 3.Solve the equation To solve the equation, we proceed as follows. x(x – 5) = 84 x 2 – 5x = 84 x 2 – 5x – 84 = 0 (x – 12)(x + 7) = 0 x – 12 = 0 or x + 7 = 0 x = 12 x = –7 Use the distributive property to remove parentheses. Subtract 84 from both sides. Factor. Apply the zero-factor property. Solve each linear equation. cont’d

7 Example We have two different values for the first integer, x = 12 or x = –7 and two different values for the second integer x – 5 = 7 or x – 5 = –12 S = {12, -7} 4.Check the result The number 7 is five less than 12 and 12  7 = 84. The number –12 is five less than –7 and –7  –12 = 84. Both pairs of integers check. cont’d

8 Solve a motion application using a quadratic equation 2.

9 Example – Flying Objects If an object is launched straight up into the air with an initial velocity of 112 feet per second, its height after t seconds is given by the formula h = 112t – 16t 2 where h represents the height of the object in feet. After this object has been launched, in how many seconds will it hit the ground?

10 Example – Flying Objects 1.What am I asked to find? Time to hit the ground: t When the object hits the ground, h = 0 2.Form an equation If we substitute 0 for h in the formula h = 112t – 16t 2, the new equation will be 0 = 112t – 16t 2 and we will solve for t. h = 112t – 16t 2 0 = 112t – 16t 2 cont’d

11 Example – Flying Objects 3.Solve the equation We solve the equation as follows. 0 = 112t – 16t 2 0 = 16t(7 – t) 16t = 0 or 7 – t = 0 t = 0 t = 7 When t = 0, the object’s height above the ground is 0 feet, because it has not been released. When t = 7, the height is again 0 feet. Factor out 16t, the GCF. Set each factor equal to 0. Solve each linear equation. cont’d

12 Example – Flying Objects 4.Check the result When t = 7, h = 112(7) – 16(7) 2 = 184 – 16(49) = 0 Since the height is 0 feet, the object has hit the ground after 7 seconds. cont’d

13 Solve a geometric application using a quadratic equation 3.

14 Example 3 – Rectangles Assume that the rectangle in Figure 5-1 has an area of 52 square centimeters and that its length is 1 centimeter more than 3 times its width. Find the perimeter of the rectangle. Figure 5-1

15 Example 3 – Rectangles 1.What am I asked to find? Perimeter of a rectangle: 2l + 2w What do we know? l = 3w + 1 lw = 52 2.Form and solve an equation w(3w + 1) = 52 cont’d

16 Example 3 – Rectangles To find the width, we can substitute 52 for A and 3w + 1 for l in the formula A = lw and solve for w. A = lw 52 = (3w + 1)w 52 = 3w 2 + w 0 = 3w 2 + w – 52 0 = (3w + 13)(w – 4) 3w + 13 = 0 or w – 4 = 0 Use the distributive property to remove parentheses. Subtract 52 from both sides. Factor. Apply the zero-factor property. cont’d

17 Example 3 – Rectangles 3w = –13 w = 4 w = Because the width of a rectangle cannot be negative, we discard the result w =. Thus, the width of the rectangle is 4 centimeters, and the length is given by 3w + 1 = 3(4) + 1 = 12 + 1 = 13 Solve each linear equation. cont’d

18 Example 3 – Rectangles The dimensions of the rectangle are 4 centimeters by 13 centimeters. We find the perimeter by substituting 13 for l and 4 for w in the formula for the perimeter. P = 2l + 2w = 2(13) + 2(4) = 26 + 8 = 34 cont’d

19 Example 3 – Rectangles 4.Check the result A rectangle with dimensions of 13 centimeters by 4 centimeters does have an area of 52 square centimeters, and the length is 1 centimeter more than 3 times the width. A rectangle with these dimensions has a perimeter of 34 centimeters. cont’d