Using the Quadratic Formula In this lesson you will learn how to use the quadratic formula to solve any quadratic equation. 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.

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

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

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

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

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 OBJECT IS THROWN: h = –16 t 2 + v t + s 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.

Modeling Vertical Motion BALLOON COMPETITION 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? SOLUTION 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.

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

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