1. In this lesson you will learn how to use the quadratic formula to solve any quadratic equation. Using the Quadratic Formula THE QUADRATIC FORMULA The solutions of the quadratic equation a x 2 + b x + c = 0 are – b  b 2 – 4 a c 2a x = when a  0 and b 2 – 4 a c ≥ 0. You can read this formula as: x equals the opposite of b, plus or minus the square root of b squared minus 4 a c, all divided by 2 a.

2. Solve x 2 + 9x + 14 = 0. Use the quadratic formula. 1x 2 + 9x + 14 = 0 – b  b 2 – 4( a )( c ) 2( a ) x = –9  81 – 56 2 x = –9  25 2 x = Identify a = 1, b = 9, and c = 14. Substitute values in the quadratic formula. Simplify. The equation has two solutions: –9  5 2 x = 9 9 1 14 1 Simplify. –9 + 5 2 x = –9 – 5 2 x = = –2 = –7 SOLUTION Using the Quadratic Formula

3. Check the solutions to x 2 + 9x + 14 = 0 in the original equation. Check x = –2 : Check x = –7 : x 2 + 9x + 14 = 0 0 = 0 (–2) 2 + 9(–2) + 14 = 0 ? 4 + –18 + 14 = 0 ? x 2 + 9x + 14 = 0 0 = 0 (–7) 2 + 9(–7) + 14 = 0 ? 49 + –63 + 14 = 0 ? Using the Quadratic Formula

4. Find the x -intercepts of the graph of y = –x 2 – 2 x + 5. Finding the x-Intercepts of a Graph The x -intercepts occur when y = 0. 0 = –1x 2 – 2 x + 5 –( b )  ( b ) 2 – 4( a )( c ) 2( a ) x = 2  4 + 20 –2 x = 2  24 –2 x = Substitute 0 for y, and identify a = 1, b = –2, and c = 5. Substitute values in the quadratic formula. Simplify. Solutions The equation has two solutions: 2  24 –2 x = 2 + 24 –2 x =  –3.45 2 – 24 –2 x =  1.45 –2 –2 –1 5 –1 SOLUTION y = –x 2 – 2 x + 5 Write original equation.

5. Check your solutions to the equation y = –x 2 – 2 x + 5 graphically. Finding the x-Intercepts of a Graph You can see that the graph shows the x -intercepts between –3 and –4 and between 1 and 2. Check y  –3.45 and y  1.45.

6. Using Quadratic Models in Real Life Problems involving models for the dropping or throwing of an object are called vertical motion problems. VERTICAL MOTION MODELS OBJECT IS DROPPED: h = –16 t 2 + s h = –16 t 2 + v t + s OBJECT IS THROWN: h = height (feet) t = time in motion (seconds) s = initial height (feet) v = initial velocity (feet per second) In these models the coefficient of t 2 is one half the acceleration due to gravity. On the surface of Earth, this acceleration is approximately 32 feet per second per second. Remember that velocity v can be positive (object moving up), negative (object moving down), or zero (object not moving). Speed is the absolute value of velocity.

7. BALLOON COMPETITION Because the marker is thrown down, the initial velocity is v = –30 feet per second. The initial height is s = 200 feet. The marker will hit the target when the height is 0. v = –30, s = 200, h = 0. SOLUTION Modeling Vertical Motion You are competing in the Field Target Event at a hot-air balloon festival. You throw a marker down from an altitude of 200 feet toward a target. When the marker leaves your hand, its speed is 30 feet per second. How long will it take the marker to hit the target?

8. BALLOON COMPETITION v = –30, s = 200, h = 0. h = –16 t 2 + v t + s h = –16 t 2 – 30t + 200 t  2.72 or –4.60 –( b )  ( b ) 2 – 4( a )( c ) 2( a ) t = 30  13,700 –32 t = –30 –30 –16 200 –16 Substitute values in the quadratic formula. Substitute 0 for h. Write in standard form. Substitute values for v and s into the vertical motion model. Choose the vertical motion model for a thrown object. Simplify. Solutions (–30) 200 0 SOLUTION Modeling Vertical Motion

9. BALLOON COMPETITION t  2.72 or –4.60 As a solution, –4.60 does not make sense in the context of the problem. SOLUTION Modeling Vertical Motion Therefore, the weighted marker will hit the target about 2.72 seconds after it was thrown.