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Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Quadratic Equations CHAPTER 11.1Completing the Square 11.2Solving Quadratic Equations Using the Quadratic Formula 11.3Solve Equations That Are Quadratic in Form 11.4Graphing Quadratic Functions 11.5Solving Nonlinear Inequalities 11

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Completing the Square Use the square root principle to solve quadratic equations. 2.Solve quadratic equations by completing the square.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Square Root Principle If x 2 = a, where a is a real number, then It is common to indicate the positive and negative solutions by writing Note: The expression is read “plus or minus the square root of a.”

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve. Solution Check Use the square root principle. Simplify by factoring out a perfect square. Check True. Note: Remember the ± means that the two solutions are and. True.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve. Solution Subtract 14 from both sides to isolate x 2. Use the square root principle. Simplify by factoring out a perfect square.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To solve equations in the form ax 2 = b, we first isolate x 2 by dividing both sides of the equation by a. We solve an equation in the form ax 2 + b = c by using both the addition and multiplication principles of equality to isolate x 2 before using the square root principle. In an equation in the form (ax + b) 2 = c, notice the expression ax + b is squared. We can use the square root principle to eliminate the square.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve. Solution Add 3 to both sides and divide each side by 5, to isolate x. Use the square root principle. or

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve by completing the square. a. Solution We first write the equation in the form x 2 + bx = c. Complete the square by adding 25 to both sides. Add 16 to both sides to get the form x 2 + bx = c. Factor. Use the square root principle. Subtract 5 from both sides to isolate x.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued b. Solution Complete the square by adding 36 to both sides. Subtract 9 from both sides to get the form x 2 + bx = c. Factor. Use the square root principle. Add 6 to both sides to isolate x. Simplify the square root.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Quadratic Equations by Completing the Square To solve a quadratic equation by completing the square: 1. Write the equation in the form x 2 + bx = c. 2. Complete the square by adding (b/2) 2 to both sides. 3. Write the completed square in factored form. 4. Use the square root principle to eliminate the square. 5. Isolate the variable. 6. Simplify as needed.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve by completing the square. Solution Divide both sides by 2. Simplify. Add to both sides to complete the square.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Combine the fractions. Factor. Add to both sides and simplify the square root. Use the square root principle.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve by completing the square. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve by completing the square. a) b) c) d)

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Quadratic Equations Using the Quadratic Formula Solve quadratic equations using the quadratic formula. 2.Use the discriminant to determine the number of real solutions that a quadratic equation will have. 3.Find the x - and y- intercepts of a quadratic function. 4.Solve applications using the quadratic formula.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Using the Quadratic Formula To solve a quadratic equation in the form ax 2 + bx + c = 0, where a 0, use the quadratic formula:

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve. Solution The equation is in the form ax 2 + bx + c = 0, where a = 2, b = –7, and c = 3.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve. Solution First we need to write the equation in the form ax 2 + bx + c = 0. a = 1, b = –2, c = –11.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley MethodWhen the Method is Beneficial 1. Factoring (Section 6.4)Use when the quadratic equation can easily be factored. 2. Square root principle (Section 9.1) Use when the quadratic equation can easily be written in the form 3. Completing the square (Section 9.1) Rarely the best method. 4. Quadratic formula (Section 9.2) Use when factoring is not easy. Methods for Solving Quadratic Equations

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Discriminant: The radicand b 2 – 4ac in the quadratic formula. We use the discriminant to determine the number and type of solutions to a quadratic equation.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Using the Discriminant Given a quadratic equation in the form ax 2 + bx + c = 0, where a 0, to determine the number and type of solutions it has, evaluate the discriminant b 2 – 4ac. If the discriminant is positive, then the equation has two real-number solutions. They will be rational if the discriminant is a perfect square and irrational otherwise. If the discriminant is 0, then the equation has one real solution. If the discriminant is negative, then the equation has two nonreal complex solutions.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Use the discriminant to determine the number and type of solutions. Solution First we write the equation in the form ax 2 + bx + c = 0. Now evaluate the discriminant, b 2 – 4ac. Replace a with 2, b with 5, and c with 1. Because the discriminant is positive, this equation has two real-number solutions.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Use the discriminant to determine the number and type of solutions. Solution Now evaluate the discriminant, b 2 – 4ac. Replace a with 1, b with –6, and c with 9. Because the discriminant is zero, this equation has only one real solution.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Use the discriminant to determine the number and type of solutions. Solution Now evaluate the discriminant, b 2 – 4ac. Replace a with 0.3, b with –0.4, and c with 0.8. Because the discriminant is negative, there are two nonreal complex solutions.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find the x- and y-intercepts of y = x 2 + x – 12, then graph. Solution Find the x-intercept by letting y = 0. 0 = x 2 + x – 12 0 = (x + 4)(x – 3) x + 4 = 0 or x – 3 = 0 x = –4 or x = 3 x-intercepts are (  4, 0) and (3, 0) Find the y-intercept, letting x = 0. y = (0) – 12 = –12 y-intercept is (0,  12)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley In physics, the general formula for describing the height of an object after it has been thrown upwards is where g represents the acceleration due to gravity, t is the time in flight, v 0 is the initial velocity, and h 0 is the initial height. For Earth, the acceleration due to gravity is  32.2 ft./sec. 2 or  9.8 m/sec. 2

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example In an extreme games competition, a motorcyclist jumps with an initial velocity of 80 feet per second from a ramp height of 25 feet, landing on a ramp with a height of 15 feet. Find the time the motorcyclist is in the air. Understand We are given the initial velocity, initial height, and final height and we are to find the time the bike is in the air. Plan Use the formula with h = 15, v 0 = 80, and h 0 = 25.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued h = 15, v 0 = 80, and h 0 = 25 Use the quadratic formula a =  16.1, b = 80 and c = 10.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Approximate the two irrational solutions. Answer Since the time cannot be negative, the motorcycle is in the air approximately 5.09 seconds. Check We can check by evaluating the original formula using t = 5.09 to see if the motorcycle lands at a height of 15 feet. We will leave this check to the reader.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve using the quadratic formula. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve using the quadratic formula. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Use the discriminant to determine the number of real solutions for the equation. a) Two real-number solutions; rational. b) Two real-number solutions; irrational. c) One real-number solution. d) No real-number solutions.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Use the discriminant to determine the number of real solutions for the equation. a) Two real-number solutions; rational. b) Two real-number solutions; irrational. c) One real-number solution. d) No real-number solutions.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Use the discriminant to determine the number of real solutions for the equation. a) Two real-number solutions; rational. b) Two real-number solutions; irrational. c) One real-number solution. d) Two nonreal complex solutions.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Use the discriminant to determine the number of real solutions for the equation. a) Two real-number solutions; rational. b) Two real-number solutions; irrational. c) One real-number solution. d) Two nonreal complex solutions.

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Equations That Are Quadratic in Form Solve equations by rewriting them in quadratic form. 2.Solve equations that are quadratic in form by using substitution. 3.Solve application problems using equations that are quadratic in form.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley An equation is quadratic in form if it can be rewritten as a quadratic equation au 2 + bu + c = 0, where a  0 and u is a variable or an expression.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve. Solution

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Check x = 10 x = –12 The solutions are –12 and 10.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve. Solution The solution is 3/4 (5 is extraneous).

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Using Substitution to Solve Equations That Are Quadratic in Form To solve equations that are quadratic in form using substitution, 1. Rewrite the equation so that it is in the form au 2 + bu + c = Solve the quadratic equation for u. 3. Substitute for u and solve. 4. Check the solutions.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solution Solve x 4 – 5x = 0. u 2 – 5u + 4 = 0 Let u = x 2. Then we solve by substituting u for x 2 and u 2 for x 4 : (u – 1)(u – 4) = 0 u = 1 or u = 4 u – 1 = 0 or u – 4 = 0 Factor Use zero factor theorem x 2 = 1 or x 2 = 4 Substitute x 2 for u. Solve by taking the square root of both sides. The solution set is {1,  1, 2,  2}.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solution Solve u 2 – 8u – 9 = 0 (u – 9)(u +1) = 0 u = 9 or u = –1 u – 9 = 0 or u + 1 = 0 Let u =. Then we solve by substituting u for and u 2 for x:

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Check: x = 81:x = 1: The solution is {81}. FALSE TRUE

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A water tower has two drainpipes attached to it. Working along, the smaller pipe would take 20 minutes longer than the larger pipe to empty the tower. If both drainpipes work together, the tower can be drained in 40 minutes. How long would it take the small pipe, working alone, to drain the tower? (Round your answer to the nearest tenth of a minute.) Solution PipeTime to Complete the Job Alone Work Rate Time Working Portion of Job Completed Smallert + 20 min.40 Largert min.40

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Use the quadratic formula.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Omit the negative solution. The amount of time required by the small pipe is represented by t + 20, it would take approximately or 91.2 minutes.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the equation. a) 2, 6 b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the equation. a) 2, 6 b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the equation. a)  9i b) 3i c)  3i d)  3

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the equation. a)  9i b) 3i c)  3i d)  3

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the equation. a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the equation. a) b) c) d)

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphing Quadratic Function Graph quadratic functions of the form f ( x ) = ax 2. 2.Graph quadratic functions of the form f ( x ) = ax 2 + k. 3.Graph quadratic functions of the form f ( x ) = a ( x – h ) 2. 4.Graph quadratic functions of the form f ( x ) = a ( x – h ) 2 + k. 5.Graph quadratic functions of the form f ( x ) = ax 2 + bx + c. 6.Solve applications involving parabolas.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Axis of Symmetry: A line that divides a graph into two symmetrical halves. Vertex: The lowest point on a parabola that opens up or highest point on a parabola that opens down.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solution xy(x, y) 0 1 –1 2 – (0, 0) (1, 2) (–1, 2) (2, 8) (–2, 8) Graph. The graph opens up because a is positive.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph. Solution xy(x, y) 0 1 –1 2 –2 0 –3 –12 (0, 0) (1, –3) (–1, –3) (2, –12) (–2, –12) The graph opens down because a is negative.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley As the |a| increases, the parabolas appear narrower.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Conclusion Given a function in the form f(x) = ax 2, the axis of symmetry is x = 0 (the y-axis) and the vertex is (0, 0). Also the greater the absolute value of a, the narrower the parabola appears (or, the smaller the absolute value of a, the wider the parabola appears).

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph g(x) = 2x 2 – 3 and h(x) = 2x Solution We will graph both functions on the same grid. The vertex of the basic function f(x) = 2x 2 is the origin (0, 0). The vertex of g(x) = 2x 2 – 3 is shifted down 3 units and the vertex of h(x) = 2x is shifted up 3 units from the origin.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Conclusion Given a function in the form f(x) = ax 2 + k, if k > 0, then the graph of f(x) = ax 2 is shifted k units up from the origin. If k < 0, then the graph of f(x) = ax 2 is shifted k units down from the origin. The new position of the vertex is (0, k). The axis of symmetry is x = 0 (the y-axis).

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph: f(x) = 3(x – 3) 2 and n(x) = 3(x + 3) 2 Solution The vertex of the basic function f(x) = 3x 2 is (0, 0) and the axis of symmetry is x = 0. The graph of f(x) = 3(x – 3) 2 has the same shape, only the vertex and axis of symmetry are shifted right 3 units from the origin along the x-axis so that they are (3, 0) and x = 3. Similarly, the vertex of n(x) = 3(x + 3) 2 is shifted 3 units to the left.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Conclusion Given a function in the form f(x) = a(x – h) 2, if h > 0, then the graph of f(x) = ax 2 is shifted h units right from the origin. If h < 0, then the graph of f(x) = ax 2 is shifted h units left from the origin. The new position of the vertex is (h, 0). The axis of symmetry is x = h.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Parabola with Vertex (h, k) The graph of a function in the form f(x) = a(x – h )2 + k is a parabola with vertex at (h, k). The equation of the axis of symmetry is x = h. The parabola opens upwards if a > 0 and downwards if a < 0. The larger the |a|, the narrower the graph. (h, k) a > 0 a < 0 x = h

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Given determine whether the graph opens up or down, find the vertex and axis of symmetry, and draw the graph. Solution a = 2, h = 2, and k =  3. a is positive the parabola opens upwards. The vertex is at (2,  3) and the axis of symmetry is x = 2.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Write f(x) = x 2 – 6x + 8 in the form f(x) = a(x – h) 2 + k. Then determine whether the graph opens upwards or downwards, find the vertex and axis of symmetry, and draw the graph. Solution Complete the square. y = x 2 – 6x + 8 y – 8 = x 2 – 6x y – = x 2 – 6x + 9 y + 1 = (x – 3) 2 y = (x – 3) 2 – 1 Since a = 1 and 1 is positive, the graph opens upwards. The vertex is at (3,  1).

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued y = x 2 – 6x + 8 xy(x, y) –1 0 8 (0, 8) (2, 0) (3, –1) (4, 0) (6, 8) y-intercept x-intercept vertex

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Finding the Vertex of a Quadratic Function in the Form f(x) = ax 2 + bx + c Given an equation in the form y = ax 2 + bx + c, to determine the vertex, 1. Find the x-coordinate using the formula 2. Find the y-coordinate by evaluating

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example For the function f(x) = 3x 2 – 12x + 4, find the coordinates of the vertex. Solution The vertex is (2,  8).

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A toy rocket is launched straight up with an initial velocity of 60 feet per second. The equation h =  16t t describes the height, h, of the rocket, t seconds after being launched. a. Find the maximum height that the rocket reaches. b. Find the amount of time that the rocket is in the air.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Since the graph of h =  16t t is a parabola that opens down, the maximum height occurs at its vertex. Vertex: h =  16(1.875) (1.875) = Vertex is at (1.875, 56.25) The maximum height is feet, which occurs seconds after the rocket is launched.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued The time the rocket is in the air is from launch until it returns to the ground. At launch and upon returning to the ground, the rocket’s height is 0, so we need to find t when h = 0. 0 =  16t t 0 =  4t(4t  15)  4t = 0 or 4t  15 = t = 0 t = 3.75 The height is 0 when t = 0 and when t = 3.75 seconds, so the rocket is in the air for 3.75 seconds.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What are the coordinates of the vertex of the function y = x 2 + 4x + 5? a) (  1, 2) b) (0, 4) c) (  2, 1) d) (4, 0)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What are the coordinates of the vertex of the function y = x 2 + 4x + 5? a) (  1, 2) b) (0, 4) c) (  2, 1) d) (4, 0)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What is the vertex of y = –2(x + 3) 2 ? a) (–3,0) b) (3,0) c) (0,–3) d) (0,3)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What is the vertex of y = –2(x + 3) 2 ? a) (–3,0) b) (3,0) c) (0,–3) d) (0,3)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What is the line of symmetry for the equation y = –2(x + 3) 2 ? a) x = 3 b) x = –3 c) y = 3 d) y = –3

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What is the line of symmetry for the equation y = –2(x + 3) 2 ? a) x = 3 b) x = –3 c) y = 3 d) y = –3

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Nonlinear Inequalities Solve quadratic and other inequalities. 2.Solve rational inequalities.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Quadratic inequality: An inequality that can be written in the form ax 2 + bx + c > 0 or ax 2 + bx + c < 0, where a  0.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Quadratic Inequalities 1. Solve the related equation ax 2 + bx + c = Plot the solutions of ax 2 + bx + c = 0 on a number line. These solutions will divide the number line into intervals. 3. Choose a test number from each interval and substitute the number into the inequality. If the test number makes the inequality true, then all numbers in that interval will solve the inequality. If the test number makes the inequality false, then no numbers in that interval will solve the inequality. 4. State the solution set of the inequality: It is the union of all the intervals that solve the inequality. If the inequality symbols are  or , then the values from step 2 are included. If the symbols are, they are not solutions.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve x 2 + x  12. Write the solution set using interval notation, then graph the solution set on a number line. Solution We begin by writing the related equation. x 2 + x – 12 = 0 (x – 3)(x + 4) = 0 x = 3 or x = –4 Plot the points on a number line and create boundaries between the three intervals IIIIII Set the quadratic expression equal to 0. Factor. Use the zero-factor theorem.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Choose a test number from each interval and substitute that value into x 2 + x – 12  IIIIII Test Point (x) For ( ,  4], we choose  5 For [  4, 3] we choose 0 For [3,  ) we choose 4 (–5) 2 + (–5) – 12 = 25 – 5 – 12 = 8 False (0) 2 + (0) – 12 = 0 – 12 = –12 True (4) 2 + (4) – 12 = – 12 = 8 False The solution is the interval [  4, 3] [ ]

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve x 2 + 5x > 0. Write the solution set using interval notation, then graph the solution set. Solution Write the related equation In interval notation the solution is For ( ,  5), we choose  6 For (  5, 0) we choose  1 For (0,  ) we choose 1 (–6) 2 + 5(–6) > 0 36 – 30 > 0 6 > 0 True (–1) 2 + 5(–1) > 0 1 – 5 > 0 –4 > 0 False (1) 2 + 5(1) > > 0 6 > 0 True ) (

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve (x + 3)(x + 1)(x – 2) < 0. Write the solution set in interval notation, then graph the solution set on a number line. Solution (x + 3)(x + 1)(x – 2) = 0 Write the related equation. x + 3 = 0 or x + 1 = 0 or x – 2 = 0 Set each factor equal to 0. x = –3 x = –1 x = 2 Solve the equations. Plot on a number line and note the intervals

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued (x + 3)(x + 1)(x – 2) < 0 Therefore, the solution set is ( ,  3)  (  1, 2) Interval ( ,  3)(  3,  1)(  1, 2)(2,  ) Test Number 44 22 03 Test Results  18 < 0 4 < 0  6 < 0 24 < 0 True or FalseTrueFalseTrueFalse ))(

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rational Inequality: An inequality containing a rational expression.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Rational Inequalities 1. Find all values that make any denominator equal to 0. These values must be excluded from the solution set. 2. Solve the related equation. 3. Plot the numbers found in steps 1 and 2 on a number line. 4.Choose a test number from each interval to determine whether it solves the inequality. 5. The solution set is the union of all the regions whose test number solves the inequality. If the inequality symbol is  or , include the values found in step 1. The solution set never includes the values found in step 2 because they make the denominator equal to 0.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve. Write the solution set using interval notation, then graph the solution set on a number line. Solution Find the values that make the denominator equal to 0. x – 2 = 0 x = 2

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Now, solve the related equation. Plot on a number line and note the intervals

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued Use a table The solution in interval notation is [  6,  4]  (2,  ). Interval ( ,  6][  6,  4][  4, 2)(2,  ) Test Number 77 55 03 Test Results  1/3   0  12  063  0 True or FalseFalseTrueFalseTrue [ ] (

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Which graph is the solution set to the inequality (x – 9)(x + 3) > 0? a) b) c) d) ) ( ( )( ) (

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Which graph is the solution set to the inequality (x – 9)(x + 3) > 0? a) b) c) d) ) ( ( )( ) (

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the inequality. y 2 + 2y + 14  0. a) ( ,  ) b)  c) [  14, 14] d) ( , 0)  (0,  )

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the inequality. y 2 + 2y + 14  0. a) ( ,  ) b)  c) [  14, 14] d) ( , 0)  (0,  )

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the inequality a) b) c) d)

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve the inequality a) b) c) d)