Copyright 2006 Pearson Education Inc Publishing as Pearson
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 8 - 1
8 8. 1 8. 2 8. 3 8. 4 8. 5 8. 6 8. 7 8. 8 8. 9 Quadratic Functions and Equations Quadratic Equations The Quadratic Formula Applications Involving Quadratic Equations Studying Solutions of Quadratic Equations Reducible to Quadratic Functions and Their Graphs More About Graphing Quadratic Functions Problem Solving and Quadratic Functions Polynomial and Rational Inequalities Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
8. 4 Studying Solutions of Quadratic Equations n The Discriminant n Writing Equations from Solutions Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
The Discriminant It is sometimes enough to know what type of number a solution will be, without actually solving the equation. From the quadratic formula, b 2 – 4 ac, is known as the discriminant. The discriminant determines what type of number the solutions of a quadratic equation are. The cases are summarized on the next slide. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 8 - 4
Discriminant b 2 – 4 ac 0 Nature of Solutions One solution; a rational number Positive Two different real-number solutions Perfect square Solutions are rational Not a perfect square Solutions are irrational conjugates Negative Two different imaginary-number solutions (complex conjugates) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 8 - 5
Example For the equation 4 x 2 – x + 1 = 0, determine what type of number the solutions are and how many exist. Solution First determine a, b, and c: a = 4, b = – 1, and c = 1. Compute the discriminant: b 2 – 4 ac = (– 1)2 – 4(4)(1) = – 15. Since the discriminant is negative, there are two imaginary-number solutions that are complex conjugates of each other. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 8 - 6
Example For the equation 5 x 2 – 10 x + 5 = 0, determine what type of number the solutions are and how many exist. Solution First determine a, b, and c: a = 5, b = – 10, and c = 5. Compute the discriminant: b 2 – 4 ac = (– 10)2 – 4(5)(5) = 0. There is exactly one solution, and it is rational. This indicates that 5 x 2 – 10 x + 5 = 0 can be solved by factoring. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 8 - 7
Example For the equation 2 x 2 + 7 x – 3 = 0, determine what type of number the solutions are and how many exist. Solution First determine a, b, and c: a = 2, b = 7, and c = – 3. Compute the discriminant: b 2 – 4 ac = (7)2 – 4(2)(– 3) = 73. The discriminant is a positive number that is not a perfect square. Thus there are two irrational solutions that are conjugates of each other. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 8 - 8
Writing Equations from Solutions We know by the principle of zero products that (x – 1)(x + 4) = 0 has solutions 1 and -4. If we know the solutions of an equation, we can write an equation, using the principle in reverse. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 8 - 9
Example Find an equation for which 5 and – 4/3 are solutions. Solution x = 5 or x = – 4/3 x – 5 = 0 or x + 4/3 = 0 (x – 5)(x + 4/3) = 0 x 2 – 5 x + 4/3 x – 20/3 = 0 3 x 2 – 11 x – 20 = 0 Get 0’s on one side Using the principle of zero products Multiplying Combining like terms and clearing fractions Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 8 - 10
Example Find an equation for which 3 i and – 3 i are solutions. Solution x = 3 i or x = – 3 i x – 3 i = 0 or x + 3 i = 0 Get 0’s on one side (x – 3 i)(x + 3 i) = 0 Using the principle of zero products x 2 – 3 ix + 3 ix – 9 i 2 = 0 x 2 + 9 = 0 Multiplying Combining like terms Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 8 - 11
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