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21 solutions of 2nd degree equations
- 1. Real Solutions of2nd Degree Equations
- 2. Real Solutions of2nd Degree Equations Recall for 3x – 5 = 0 its solution is x = –(–5)/3 = 5/3
- 3. Real Solutions of2nd Degree Equations Recall for 3x – 5 = 0 its solution is x = –(–5)/3 = 5/3 and that in general for a given 1st degree or linear equation of the form ax + b = 0, the solution is x = –b/a.
- 4. Real Solutions of2nd Degree Equations Recall for 3x – 5 = 0 its solution is x = –(–5)/3 = 5/3 and that in general for a given 1st degree or linear equation of the form ax + b = 0, the solution is x = –b/a. Note that in this case, the solution is given as a “simple” formula in terms of the coefficients a and b.
- 5. Real Solutions of2nd Degree Equations Recall for 3x – 5 = 0 its solution is x = –(–5)/3 = 5/3 and that in general for a given 1st degree or linear equation of the form ax + b = 0, the solution is x = –b/a. Note that in this case, the solution is given as a “simple” formula in terms of the coefficients a and b. A 2nd degree or quadratic equations is an equation of the form ax2 + bx + c = 0.
- 6. Real Solutions of2nd Degree Equations Recall for 3x – 5 = 0 its solution is x = –(–5)/3 = 5/3 and that in general for a given 1st degree or linear equation of the form ax + b = 0, the solution is x = –b/a. Note that in this case, the solution is given as a “simple” formula in terms of the coefficients a and b. A 2nd degree or quadratic equations is an equation of the form ax2 + bx + c = 0. Its two solutions may also be given in terms of –b±√ b2 – 4ac 2aQuadratic Formula where x = .a, b and c by the
- 7. Real Solutions of2nd Degree Equations Recall for 3x – 5 = 0 its solution is x = –(–5)/3 = 5/3 and that in general for a given 1st degree or linear equation of the form ax + b = 0, the solution is x = –b/a. Note that in this case, the solution is given as a “simple” formula in terms of the coefficients a and b. A 2nd degree or quadratic equations is an equation of the form ax2 + bx + c = 0. Its two solutions may also be given in terms of The verifications of this formula is based on an exclusive 2nd –deg–method called completing the square. –b±√ b2 – 4ac 2aQuadratic Formula where x = .a, b and c by the
- 8. Real Solutions of2nd Degree Equations Recall for 3x – 5 = 0 its solution is x = –(–5)/3 = 5/3 and that in general for a given 1st degree or linear equation of the form ax + b = 0, the solution is x = –b/a. Note that in this case, the solution is given as a “simple” formula in terms of the coefficients a and b. A 2nd degree or quadratic equations is an equation of the form ax2 + bx + c = 0. Its two solutions may also be given in terms of The verifications of this formula is based on an exclusive 2nd –deg–method called completing the square. We will study this method and present the derivation of the quadratic formula later. –b±√ b2 – 4ac 2aQuadratic Formula where x = .a, b and c by the
- 9. Real Solutions of2nd Degree Equations Recall for 3x – 5 = 0 its solution is x = –(–5)/3 = 5/3 and that in general for a given 1st degree or linear equation of the form ax + b = 0, the solution is x = –b/a. Note that in this case, the solution is given as a “simple” formula in terms of the coefficients a and b. A 2nd degree or quadratic equations is an equation of the form ax2 + bx + c = 0. Its two solutions may also be given in terms of The verifications of this formula is based on an exclusive 2nd –deg–method called completing the square. We will study this method and present the derivation of the quadratic formula later. –b±√ b2 – 4ac 2aQuadratic Formula where x = .a, b and c by the Let’s review the standard methods for finding real number solutions of ax2 + bx + c = 0 below.
- 10. Real Solutions of2nd Degree Equations To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing.
- 11. Real Solutions of2nd Degree Equations To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing.
- 12. Real Solutions of2nd Degree Equations To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials.
- 13. Real Solutions of2nd Degree Equations To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Quadratic Formula (QF).
- 14. Real Solutions of2nd Degree Equations The Square-Root Method To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Quadratic Formula (QF).
- 15. Real Solutions of2nd Degree Equations The Square-Root Method Use the square-root method if there is no x-term. To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Quadratic Formula (QF).
- 16. Real Solutions of2nd Degree Equations The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Quadratic Formula (QF).
- 17. Real Solutions of2nd Degree Equations The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = ±√d are the roots. To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Quadratic Formula (QF).
- 18. Real Solutions of2nd Degree Equations The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = ±√d are the roots. Example A. Solve 3x2 – 7 = 0 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Quadratic Formula (QF).
- 19. Real Solutions of2nd Degree Equations The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = ±√d are the roots. Example A. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Quadratic Formula (QF).
- 20. Real Solutions of2nd Degree Equations The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = ±√d are the roots. Example A. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = 7 3 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Quadratic Formula (QF).
- 21. Real Solutions of2nd Degree Equations The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = ±√d are the roots. Example A. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = ±√7/3 7 3 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Quadratic Formula (QF).
- 22. Real Solutions of2nd Degree Equations The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d II. Then x = ±√d are the roots. Example A. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = ±√7/3 ≈ ±1.53 7 3 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Quadratic Formula (QF).
- 23. Real Solutions of2nd Degree Equations The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d II. Then x = ±√d are the roots. Example A. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = ±√7/3 ≈ ±1.53 7 3 exact answers To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Quadratic Formula (QF).
- 24. Real Solutions of2nd Degree Equations The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d II. Then x = ±√d are the roots. Example A. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = ±√7/3 ≈ ±1.53 7 3 exact answers approx. answers To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Quadratic Formula (QF).
- 25. Factoring Method Real Solutionsof 2nd Degree Equations
- 26. Factoring Method This isthe method we learned in beginning algebra. Real Solutions of 2nd Degree Equations
- 27. Factoring Method This isthe method we learned in beginning algebra. For example, to solve x2 – x = 2 Real Solutions of 2nd Degree Equations
- 28. Factoring Method This isthe method we learned in beginning algebra. For example, to solve x2 – x = 2 set one side to 0, x2 – x – 2 = 0, Real Solutions of 2nd Degree Equations
- 29. Factoring Method This isthe method we learned in beginning algebra. For example, to solve x2 – x = 2 set one side to 0, x2 – x – 2 = 0, factor, (x – 2)(x + 1) = 0, Real Solutions of 2nd Degree Equations
- 30. Factoring Method This isthe method we learned in beginning algebra. For example, to solve x2 – x = 2 set one side to 0, x2 – x – 2 = 0, factor, (x – 2)(x + 1) = 0, then set each binomial to 0, x – 2 = 0 or that x + 1 = 0, Real Solutions of 2nd Degree Equations
- 31. Factoring Method This isthe method we learned in beginning algebra. For example, to solve x2 – x = 2 set one side to 0, x2 – x – 2 = 0, factor, (x – 2)(x + 1) = 0, then set each binomial to 0, x – 2 = 0 or that x + 1 = 0, and we obtain that x = 2 or –1. Real Solutions of 2nd Degree Equations
- 32. Factoring Method This isthe method we learned in beginning algebra. For example, to solve x2 – x = 2 set one side to 0, x2 – x – 2 = 0, factor, (x – 2)(x + 1) = 0, then set each binomial to 0, x – 2 = 0 or that x + 1 = 0, and we obtain that x = 2 or –1. We summarize the steps for the factoring method here. I. Factor the equation as (#x + #)(#x + #) = 0 II. Extract the answers by setting each binomial to be 0. Real Solutions of 2nd Degree Equations
- 33. Factoring Method This isthe method we learned in beginning algebra. For example, to solve x2 – x = 2 set one side to 0, x2 – x – 2 = 0, factor, (x – 2)(x + 1) = 0, then set each binomial to 0, x – 2 = 0 or that x + 1 = 0, and we obtain that x = 2 or –1. We summarize the steps for the factoring method here. I. Factor the equation as (#x + #)(#x + #) = 0 II. Extract the answers by setting each binomial to be 0. Real Solutions of 2nd Degree Equations However this method are restricted to factorable equations.
- 34. Factoring Method This isthe method we learned in beginning algebra. For example, to solve x2 – x = 2 set one side to 0, x2 – x – 2 = 0, factor, (x – 2)(x + 1) = 0, then set each binomial to 0, x – 2 = 0 or that x + 1 = 0, and we obtain that x = 2 or –1. We summarize the steps for the factoring method here. I. Factor the equation as (#x + #)(#x + #) = 0 II. Extract the answers by setting each binomial to be 0. Real Solutions of 2nd Degree Equations However this method are restricted to factorable equations. This method fails for x2 + 1 = 0 because x2 + 1 is not factorable.
- 35. Factoring Method This isthe method we learned in beginning algebra. For example, to solve x2 – x = 2 set one side to 0, x2 – x – 2 = 0, factor, (x – 2)(x + 1) = 0, then set each binomial to 0, x – 2 = 0 or that x + 1 = 0, and we obtain that x = 2 or –1. We summarize the steps for the factoring method here. I. Factor the equation as (#x + #)(#x + #) = 0 II. Extract the answers by setting each binomial to be 0. Real Solutions of 2nd Degree Equations However this method are restricted to factorable equations. This method fails for x2 + 1 = 0 because x2 + 1 is not factorable. When we attempt but fail in factoring an equation, issues arise.
- 36. Factoring Method This isthe method we learned in beginning algebra. For example, to solve x2 – x = 2 set one side to 0, x2 – x – 2 = 0, factor, (x – 2)(x + 1) = 0, then set each binomial to 0, x – 2 = 0 or that x + 1 = 0, and we obtain that x = 2 or –1. We summarize the steps for the factoring method here. I. Factor the equation as (#x + #)(#x + #) = 0 II. Extract the answers by setting each binomial to be 0. Real Solutions of 2nd Degree Equations However this method are restricted to factorable equations. This method fails for x2 + 1 = 0 because x2 + 1 is not factorable. When we attempt but fail in factoring an equation, issues arise. Is the equation factorable?
- 37. Factoring Method This isthe method we learned in beginning algebra. For example, to solve x2 – x = 2 set one side to 0, x2 – x – 2 = 0, factor, (x – 2)(x + 1) = 0, then set each binomial to 0, x – 2 = 0 or that x + 1 = 0, and we obtain that x = 2 or –1. We summarize the steps for the factoring method here. I. Factor the equation as (#x + #)(#x + #) = 0 II. Extract the answers by setting each binomial to be 0. Real Solutions of 2nd Degree Equations However this method are restricted to factorable equations. This method fails for x2 + 1 = 0 because x2 + 1 is not factorable. When we attempt but fail in factoring an equation, issues arise. Is the equation factorable? How can we confirm easily that it’s not possible to factor the equation so we may stop waiting time trying to do it ?
- 38. Factoring Method This isthe method we learned in beginning algebra. For example, to solve x2 – x = 2 set one side to 0, x2 – x – 2 = 0, factor, (x – 2)(x + 1) = 0, then set each binomial to 0, x – 2 = 0 or that x + 1 = 0, and we obtain that x = 2 or –1. We summarize the steps for the factoring method here. I. Factor the equation as (#x + #)(#x + #) = 0 II. Extract the answers by setting each binomial to be 0. Real Solutions of 2nd Degree Equations However this method are restricted to factorable equations. This method fails for x2 + 1 = 0 because x2 + 1 is not factorable. When we attempt but fail in factoring an equation, issues arise. Is the equation factorable? How can we confirm easily that it’s not possible to factor the equation so we may stop waiting time trying to do it ? What are the solutions of the equation if it’s not factorable?
- 39. Factoring Method This isthe method we learned in beginning algebra. For example, to solve x2 – x = 2 set one side to 0, x2 – x – 2 = 0, factor, (x – 2)(x + 1) = 0, then set each binomial to 0, x – 2 = 0 or that x + 1 = 0, and we obtain that x = 2 or –1. We summarize the steps for the factoring method here. I. Factor the equation as (#x + #)(#x + #) = 0 II. Extract the answers by setting each binomial to be 0. Real Solutions of 2nd Degree Equations However this method are restricted to factorable equations. This method fails for x2 + 1 = 0 because x2 + 1 is not factorable. When we attempt but fail in factoring an equation, issues arise. Is the equation factorable? How can we confirm easily that it’s not possible to factor the equation so we may stop waiting time trying to do it ? What are the solutions of the equation if it’s not factorable? The Quadratic Formula (QF) answers all these questions.
- 40. Real Solutions of2nd Degree Equations Quadratic Formula (QF)
- 41. –b±√ b2 – 4ac 2a RealSolutions of 2nd Degree Equations Quadratic Formula (QF) The roots for the equation ax2 + bx + c = 0 are x = ,
- 42. –b±√ b2 – 4ac 2a RealSolutions of 2nd Degree Equations Quadratic Formula (QF) The roots for the equation ax2 + bx + c = 0 are x = , b2 – 4ac is called the discriminant because its value indicates what type of roots there are.
- 43. –b±√ b2 – 4ac 2a RealSolutions of 2nd Degree Equations Quadratic Formula (QF) The roots for the equation ax2 + bx + c = 0 are x = , b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically if * b2 – 4ac = 0, 1, 4, 9, 16, etc.. is a perfect square,
- 44. –b±√ b2 – 4ac 2a RealSolutions of 2nd Degree Equations Quadratic Formula (QF) The roots for the equation ax2 + bx + c = 0 are x = , b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically if * b2 – 4ac = 0, 1, 4, 9, 16, etc.. is a perfect square, so √ b2 – 4ac the square root is a whole number,
- 45. –b±√ b2 – 4ac 2a RealSolutions of 2nd Degree Equations Quadratic Formula (QF) The roots for the equation ax2 + bx + c = 0 are x = , b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically if * b2 – 4ac = 0, 1, 4, 9, 16, etc.. is a perfect square, so √ b2 – 4ac the square root is a whole number, then the expression is factorable.
- 46. –b±√ b2 – 4ac 2a RealSolutions of 2nd Degree Equations Quadratic Formula (QF) The roots for the equation ax2 + bx + c = 0 are x = , b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically if * b2 – 4ac = 0, 1, 4, 9, 16, etc.. is a perfect square, so √ b2 – 4ac the square root is a whole number, then the expression is factorable. * b2 – 4ac is not a square number listed above, then its square is irrational.
- 47. –b±√ b2 – 4ac 2a RealSolutions of 2nd Degree Equations Quadratic Formula (QF) The roots for the equation ax2 + bx + c = 0 are x = , b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically if * b2 – 4ac = 0, 1, 4, 9, 16, etc.. is a perfect square, so √ b2 – 4ac the square root is a whole number, then the expression is factorable. * b2 – 4ac is not a square number listed above, then its square is irrational. In particular if b2 – 4ac < 0 there is no real roots.
- 48. –b±√ b2 – 4ac 2a RealSolutions of 2nd Degree Equations Example C. Solve 3x2 – 2x – 2 = 0 Quadratic Formula (QF) The roots for the equation ax2 + bx + c = 0 are x = , b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically if * b2 – 4ac = 0, 1, 4, 9, 16, etc.. is a perfect square, so √ b2 – 4ac the square root is a whole number, then the expression is factorable. * b2 – 4ac is not a square number listed above, then its square is irrational. In particular if b2 – 4ac < 0 there is no real roots.
- 49. –b±√ b2 – 4ac 2a RealSolutions of 2nd Degree Equations Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable, Quadratic Formula (QF) The roots for the equation ax2 + bx + c = 0 are x = , b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically if * b2 – 4ac = 0, 1, 4, 9, 16, etc.. is a perfect square, so √ b2 – 4ac the square root is a whole number, then the expression is factorable. * b2 – 4ac is not a square number listed above, then its square is irrational. In particular if b2 – 4ac < 0 there is no real roots.
- 50. –b±√ b2 – 4ac 2a RealSolutions of 2nd Degree Equations Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable, a = 3, b = –2, c = –2 hence b2 – 4ac = (–2)2 – 4(3)(–2) = 28 Quadratic Formula (QF) The roots for the equation ax2 + bx + c = 0 are x = , b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically if * b2 – 4ac = 0, 1, 4, 9, 16, etc.. is a perfect square, so √ b2 – 4ac the square root is a whole number, then the expression is factorable. * b2 – 4ac is not a square number listed above, then its square is irrational. In particular if b2 – 4ac < 0 there is no real roots.
- 51. –b±√ b2 – 4ac 2a RealSolutions of 2nd Degree Equations Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable, a = 3, b = –2, c = –2 hence b2 – 4ac = (–2)2 – 4(3)(–2) = 28 which is not a square number, so it’s not factorable! Quadratic Formula (QF) The roots for the equation ax2 + bx + c = 0 are x = , b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically if * b2 – 4ac = 0, 1, 4, 9, 16, etc.. is a perfect square, so √ b2 – 4ac the square root is a whole number, then the expression is factorable. * b2 – 4ac is not a square number listed above, then its square is irrational. In particular if b2 – 4ac < 0 there is no real roots.
- 52. –b±√ b2 – 4ac 2a RealSolutions of 2nd Degree Equations Example C. Solve 3x2 – 2x – 2 = 0 Check if it is factorable, a = 3, b = –2, c = –2 hence b2 – 4ac = (–2)2 – 4(3)(–2) = 28 which is not a square number, so it’s not factorable! By QF, x = 2 ± √ 28 6 = {1.22 –0.549 Quadratic Formula (QF) The roots for the equation ax2 + bx + c = 0 are x = , b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically if * b2 – 4ac = 0, 1, 4, 9, 16, etc.. is a perfect square, so √ b2 – 4ac the square root is a whole number, then the expression is factorable. * b2 – 4ac is not a square number listed above, then its square is irrational. In particular if b2 – 4ac < 0 there is no real roots.
- 53. Real Solutions of2nd Degree Equations Example D. Solve 3x2 – 2x + 2 = 0
- 54. Real Solutions of2nd Degree Equations Example D. Solve 3x2 – 2x + 2 = 0 To check if it is factorable: a = 3, b = –2, c = 2,
- 55. Real Solutions of2nd Degree Equations Example D. Solve 3x2 – 2x + 2 = 0 To check if it is factorable: a = 3, b = –2, c = 2, so b2 – 4ac = (–2)2 – 4(3)(2) = –20,
- 56. Real Solutions of2nd Degree Equations Example D. Solve 3x2 – 2x + 2 = 0 To check if it is factorable: a = 3, b = –2, c = 2, so b2 – 4ac = (–2)2 – 4(3)(2) = –20, √–20 is not a real number, so there is no real solution.
- 57. Real Solutions of2nd Degree Equations Example D. Solve 3x2 – 2x + 2 = 0 To check if it is factorable: a = 3, b = –2, c = 2, so b2 – 4ac = (–2)2 – 4(3)(2) = –20, √–20 is not a real number, so there is no real solution. Quadratic equations appear in problems about distances or about areas, in finance, and in all branches of social and physical sciences.
- 58. Real Solutions of2nd Degree Equations Example D. Solve 3x2 – 2x + 2 = 0 To check if it is factorable: a = 3, b = –2, c = 2, so b2 – 4ac = (–2)2 – 4(3)(2) = –20, √–20 is not a real number, so there is no real solution. Quadratic equations appear in problems about distances or about areas, in finance, and in all branches of social and physical sciences. The 2nd degree and the1st degree equations constitute the most basic and important classes of equations.
- 59. Real Solutions of2nd Degree Equations Example D. Solve 3x2 – 2x + 2 = 0 To check if it is factorable: a = 3, b = –2, c = 2, so b2 – 4ac = (–2)2 – 4(3)(2) = –20, √–20 is not a real number, so there is no real solution. Quadratic equations appear in problems about distances or about areas, in finance, and in all branches of social and physical sciences. The 2nd degree and the1st degree equations constitute the most basic and important classes of equations. Finally, are there similar solution–formulas for higher degree polynomial equations in terms of their coefficients a, b,...?
- 60. Real Solutions of2nd Degree Equations Example D. Solve 3x2 – 2x + 2 = 0 To check if it is factorable: a = 3, b = –2, c = 2, so b2 – 4ac = (–2)2 – 4(3)(2) = –20, √–20 is not a real number, so there is no real solution. Quadratic equations appear in problems about distances or about areas, in finance, and in all branches of social and physical sciences. The 2nd degree and the1st degree equations constitute the most basic and important classes of equations. Finally, are there similar solution–formulas for higher degree polynomial equations in terms of their coefficients a, b,...? It turns out that there are such solution–formulas for polynomials of degree N = 1, 2, 3, 4,
- 61. Real Solutions of2nd Degree Equations Example D. Solve 3x2 – 2x + 2 = 0 To check if it is factorable: a = 3, b = –2, c = 2, so b2 – 4ac = (–2)2 – 4(3)(2) = –20, √–20 is not a real number, so there is no real solution. Quadratic equations appear in problems about distances or about areas, in finance, and in all branches of social and physical sciences. The 2nd degree and the1st degree equations constitute the most basic and important classes of equations. Finally, are there similar solution–formulas for higher degree polynomial equations in terms of their coefficients a, b,...? It turns out that there are such solution–formulas for polynomials of degree N = 1, 2, 3, 4, but for degree N = 5, 6,.. , there do not exist solution–formulas in terms of the coefficients a, b,…
- 62. Real Solutions of2nd Degree Equations Example D. Solve 3x2 – 2x + 2 = 0 To check if it is factorable: a = 3, b = –2, c = 2, so b2 – 4ac = (–2)2 – 4(3)(2) = –20, √–20 is not a real number, so there is no real solution. Quadratic equations appear in problems about distances or about areas, in finance, and in all branches of social and physical sciences. The 2nd degree and the1st degree equations constitute the most basic and important classes of equations. Finally, are there similar solution–formulas for higher degree polynomial equations in terms of their coefficients a, b,...? It turns out that there are such solution–formulas for polynomials of degree N = 1, 2, 3, 4, but for degree N = 5, 6,.. , there do not exist solution–formulas in terms of the coefficients a, b,… Specifically, the solutions of the equation 3x5 – 2x4 + 1.7x3 – 0.2x +√2 = 0 cant be solved by simplify by plugging in the coefficients 3, –2, 1.7, –0.2 into a formula.,
- 63. Real Solutions of2nd Degree Equations 1. 3x2 – 12 = 0 Exercise A. Solve. Use the square root methods. Give both the exact answers and the numerical answers. If the answer is not real, state so. 2. 3x2 – 15 = 0 3. 3x2 + 15 = 0 4. x2 – 3 = –x2 + 15 5. –6 = 4x2 – 15 6. (x – 2)2 = 2 7. 4 = (2x – 3)2 Exercise B. Solve by factoring and by the quadratic formula. 14. x2 – 3x = 10 15. x(x – 2) = 24 16. 2x2 = 3(x + 1) – 1 8. x2 – 3x = 4 9. x – 15 = 2x 10. x2 + 5x + 12 = 0 11. –x2 – 2x + 8 = 0 12. 9 – 3x – 2x2 = 0 13. 2x2 – x – 1 = 0 Exercise C. Solve by the quadratic formula. 17. x2 – x + 1 = 0 18. x2 – x – 1 = 0 19. x2 – 3x – 2 = 0 20. x2 – 2x + 3 = 0 If the answers are not real numbers, state so. 21. 2x2 – 3x – 1 = 0 22. 3x2 = 2x + 3
- 64. Real Solutions of2nd Degree Equations Exercise C. Solve by the quadratic formula. If the answers are not real numbers, state so. 23. 2(x2 – 1) + x = 4 24. (x – 1)(x + 1) = 2x(x + 2) 25. (x – 1) (x + 2) = 2x (x + 2) 26. (x + 1) (x + 2) = (x + 2) (2x + 1) 1 L 1 – L 1 – L L1 L = 27. Cut a stick of unit length 1 into two parts in a manner such that “the ratio of the whole 1 to the large part Ф, is the same as, the ratio of the large part L to the small part 1 – L” In picture, This ratio 1 : L is the golden ratio Ф (phi). Find the exact and approximate value of Ф then google “golden ratio” and list one of its applications.