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4 th EDITION

College Algebra & Trigonometry and Precalculus. 4 th EDITION. Quadratic Equations. 1.4. Solving a Quadratic Equation Completing the Square The Quadratic Formula Solving for a Specified Variable The Discriminant. Quadratic Equation in One Variable.

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4 th EDITION

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  1. College Algebra & Trigonometry and Precalculus 4th EDITION

  2. Quadratic Equations 1.4 Solving a Quadratic Equation Completing the Square The Quadratic Formula Solving for a Specified Variable The Discriminant

  3. Quadratic Equation in One Variable An equation that can be written in the form where a, b, and c are real numbers with a ≠ 0, is a quadratic equation. The given form is called standard form.

  4. Second-degree Equation A quadratic equation is a second-degree equation. This is an equation with a squared variable term and no terms of greater degree.

  5. Zero-Factor Property If a and b are complex numbers with ab = 0, then a = 0 or b = 0 or both.

  6. USING THE ZERO-FACTOR PROPERTY Example 1 Solve Solution: Standard form Factor. Zero-factor property.

  7. USING THE ZERO-FACTOR PROPERTY Example 1 Solve Solution: Zero-factor property. Solve each equation.

  8. Square-Root Property A quadratic equation of the form x2 = k can also be solved by factoring Subtract k. Factor. Zero-factor property. Solve each equation.

  9. Square Root Property If x2 = k, then

  10. Square-Root Property That is, the solution of Both solutions are real if k > 0, and both are imaginary if k < 0 If k = 0, then this is sometimes called a double solution. If k < 0, we write the solution set as

  11. USING THE SQUARE ROOT PROPERTY Example 2 Solve each quadratic equation. a. Solution: By the square root property, the solution set is

  12. USING THE SQUARE ROOT PROPERTY Example 2 Solve each quadratic equation. b. Solution: Since the solution set of x2 =−25 is

  13. USING THE SQUARE ROOT PROPERTY Example 2 Solve each quadratic equation. c. Solution: Use a generalization of the square root property. Generalized square root property. Add 4.

  14. Solving A Quadratic Equation By Completing The Square To solve ax2 + bx + c = 0, by completing the square: Step 1 If a ≠ 1, divide both sides of the equation by a. Step 2 Rewrite the equation so that the constant term is alone on one side of the equality symbol. Step 3 Square half the coefficient of x, and add this square to both sides of the equation. Step 4 Factor the resulting trinomial as a perfect square and combine like terms on the other side. Step 5 Use the square root property to complete the solution.

  15. USING THE METHOD OF COMPLETING THE SQUARE a = 1 Example 3 Solve x2 – 4x –14 = 0 by completing the square. Solution Step 1 This step is not necessary since a = 1. Add 14 to both sides. Step 2 Step 3 add 4 to both sides. Step 4 Factor; combine terms.

  16. USING THE METHOD OF COMPLETING THE SQUARE a = 1 Example 3 Solve x2 – 4x –14 = 0 by completing the square. Solution Step 4 Factor; combine terms. Step 5 Square root property. Take both roots. Add 2. Simplify the radical. The solution set is

  17. USING THE METHOD OF COMPLETING THE SQUARE a ≠ 1 Example 4 Solve 9x2 – 12x + 9 = 0 by completing the square. Solution Divide by 9. (Step 1) Add–1. (Step 2)

  18. USING THE METHOD OF COMPLETING THE SQUARE a = 1 Example 4 Solve 9x2 – 12x + 9 = 0 by completing the square. Solution Factor, combine terms. (Step 4) Square root property

  19. USING THE METHOD OF COMPLETING THE SQUARE a = 1 Example 4 Solve 9x2 – 12x + 9 = 0 by completing the square. Solution Square root property Quotient rule for radicals Add ⅔.

  20. USING THE METHOD OF COMPLETING THE SQUARE a = 1 Example 4 Solve 9x2 – 12x + 9 = 0 by completing the square. Solution Add ⅔. The solution set is

  21. The Quadratic Formula The method of completing the square can be used to solve any quadratic equation. If we start with the general quadratic equation, ax2 + bx + c = 0, a ≠ 0, and complete the square to solve this equation for x in terms of the constants a, b, and c, the result is a general formula for solving any quadratic equation. We assume that a > 0.

  22. Quadratic Formula The solutions of the quadratic equation ax2 + bx + c = 0, where a ≠ 0, are

  23. CautionNotice that the fraction bar in the quadratic formula extends under the –b term in the numerator.

  24. USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Example 5 Solve x2 –4x =–2 Solution: Write in standard form. Here a = 1, b = –4, c = 2 Quadratic formula.

  25. USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Example 5 Solve x2 –4x =–2 Solution: Quadratic formula. The fraction bar extends under–b.

  26. USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Example 5 Solve x2 –4x =–2 Solution: The fraction bar extends under–b.

  27. USING THE QUADRATIC FORMULA (REAL SOLUTIONS) Example 5 Solve x2 –4x =–2 Solution: Factor out 2 in the numerator. Factor first, then divide. Lowest terms. The solution set is

  28. USING THE QUADRATIC FORMULA (NONREAL COMPLEX SOLUTIONS) Example 6 Solve 2x2 = x –4. Solution: Write in standard form. Quadratic formula; a = 2, b =–1, c = 4 Use parentheses and substitute carefully to avoid errors.

  29. USING THE QUADRATIC FORMULA (NONREAL COMPLEX SOLUTIONS) Example 6 Solve 2x2 = x –4. Solution: The solution set is

  30. Cubic Equation The equation x3 + 8 = 0 that follows is called a cubic equation because of the degree 3 term. Some higher-degree equations can be solved using factoring and the quadratic formula.

  31. SOLVING A CUBIC EQUATION Example 7 Solve Solution Factor as a sum of cubes. or Zero-factor property or Quadratic formula; a = 1, b = –2, c = 4

  32. SOLVING A CUBIC EQUATION Example 7 Solve Solution Simplify. Simplify the radical. Factor out 2 in the numerator.

  33. SOLVING A CUBIC EQUATION Example 7 Solve Solution Lowest terms

  34. SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve for the specified variable. Use  when taking square roots. a. Solution Goal: Isolate d, the specified variable. Multiply by 4. Divide by .

  35. SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use  when taking square roots. a. Solution Divide by . Square root property See the Note following this example.

  36. SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use  when taking square roots. a. Solution Square root property Rationalize the denominator.

  37. SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use  when taking square roots. a. Solution Rationalize the denominator. Multiply numerators; multiply denominators.

  38. SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use  when taking square roots. a. Solution Multiply numerators; multiply denominators. Simplify.

  39. Solving for a Specified Variable Note In Example 8, we took both positive and negative square roots. However, if the variable represents a distance or length in an application, we would consider only the positive square root.

  40. SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use  when taking square roots. b. Solution Write in standard form. Now use the quadratic formula to find t.

  41. SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use  when taking square roots. b. Solution a = r, b = –s, and c = –k

  42. SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA Example 8 Solve the specified variable. Use  when taking square roots. b. Solution a = r, b = –s, and c = –k Simplify.

  43. The Discriminant The Discriminant The quantity under the radical in the quadratic formula, b2 – 4ac, is called the discriminant. Discriminant

  44. The Discriminant

  45. CautionThe restriction on a, b, and c is important. For example, for the equation the discriminant is b2 – 4ac = 5 + 4 = 9, which would indicate two rational solutions if the coefficients were integers. By the quadratic formula, however, the two solutions are irrational numbers,

  46. USING THE DISCRIMINANT Example 9 Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. a. Solution For 5x2 + 2x – 4 = 0, a = 5, b = 2, and c = –4. The discriminant is b2 – 4ac = 22 – 4(5)(–4) = 84 The discriminant is positive and not a perfect square, so there are two distinct irrational solutions.

  47. USING THE DISCRIMINANT Example 9 Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. b. Solution First write the equation in standard form as x2 – 10x + 25 = 0. Thus, a = 1, b = –10, and c = 25, and b2 – 4ac =(–10 )2 – 4(1)(25) = 0 There is one distinct rational solution, a “double solution.”

  48. USING THE DISCRIMINANT Example 9 Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. c. Solution For 2x2 – x + 1 = 0, a = 2, b = –1, and c = 1, so b2 – 4ac = (–1)2 – 4(2)(1) = –7. There are two distinct nonreal complex solutions. (They are complex conjugates.)

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