4 th ESO Unit 3: Equations. Systems of equations
1. Polynomial equations Polynomial Equation is simply a polynomial that has been set equal to zero in an equation. Linear Polynomials (Degree 1) aa + b = 0 They have only a solution: x = b a Quadratic Polynomials (Degree 2) ax 2 + bb + c = 0 Solutions: x = b ± b2 4aa 2a 2
Degree 3 and up 3
Biquadratic equations The term "biquadratic equation" is a quartic equation having no odd powers, i.e., ax 4 + bx 2 + c = 0 Such equations are easy to solve, since they reduce to a quadratic equation in the variable x 2 = z and hence can be solved for z using the quadratic formula and then in terms of the original variables by taking the square roots. Exercises 1. Solve 2. Find the solutions 3. Find the solutions 4
4. Solve the following biquadratic equations 5. Solve 6. 2. Rational equations A rational equation is an equation in which one or more of the terms are algebraic fractions (unknown in the denominator). When solving these rational equations, we will eliminate the denominator of each of the terms. Example. Solve the following equation: First, I need to check the denominators: they tell me that x cannot equal zero or 2 (since these values would cause division by zero). I'll re-check at the end, to make sure any solutions I find are "valid". I could convert everything to the common denominator of 5x(x + 2) and then compare the numerators: 5
At this point, the denominators are the same. So do they really matter? Not really (other than for saying what values x can't be). At this point, the two sides of the equation will be equal as long as the numerators are equal. That is, all I really need to do now is solve the numerators: 15x (5x + 10) = x + 2 10x 10 = x + 2 9x = 12 x = 12 / 9 = 4 / 3 Since x = 4/3 won't cause any division-by-zero problems in the fractions in the original equation, then this solution is valid. x = 4 / 3 6
Exercises. 7. Solve 8. Find the solutions Example. 7
9. 3. Irrational equations Irrational equations or radical equations have the unknown value under the radical. To solve an irrational equation, follow these steps: 1. Isolate a radical in one of the two members and pass it to another member of the other terms which are also radical. 2. Square both members. 3. Solve the equation obtained. 4. Check if the solutions obtained verify the initial equation. 5. If the equation has several radicals, repeat the first two steps of the process to remove all of them. Example 1. Isolate the radical: 3. Solve the equation: 2. Square both members: 4. Verify: The equation has the solution x = 2. 8
Example Exercises 10. Find the possible solutions 11. Solve 12. Solve 9
13. Solve 14. 4. Logarithmic Equations The first type of logarithmic equation has two logs, each having the same base, set equal to each other, and you solve by setting the insides (the "arguments") equal to each other. For example: Solve log 2 (x) = log 2 (14). Since the logarithms on either side of the equation have the same base ("2", in this case), then the only way these two logs can be equal is for their arguments to be equal. In other words, the log expressions being equal says that the arguments must be equal, so I have: x = 14 And that's the solution: x = 14 Solve log b (x 2 ) = log b (2x 1). Since the bases of the logs are the same (the unknown value "b", in this case), then the insides must be equal. That is: x 2 = 2x 1 Then I can solve the log equation by solving this quadratic equation: x 2 2x + 1 = 0 (x 1)(x 1) = 0 Then the solution is x = 1. 10
Logarithms cannot have non-positive arguments, but quadratics and other equations can have negative solutions. So it is generally a good idea to check the solutions you get for log equations: log b (x 2 ) = log b (2x 1) log b ([1] 2 )?=? log b (2[1] 1) log b (1)?=? log b (2 1) log b (1) = log b (1) The value of the base of the log is irrelevant here. Each log has the same base, each log ends up with the same argument, and that argument is a positive value, so the solution "checks". Solve 2log b (x) = log b (4) + log b (x 1). All of these logs have the same base, but I can't solve yet, because I don't yet have "log equals log". So first I'll have to apply log rules: Then: 2log b (x) = log b (4) + log b (x 1) log b (x 2 ) = log b ((4)(x 1)) log b (x 2 ) = log b (4x 4) x 2 = 4x 4 x 2 4x + 4 = 0 (x 2)(x 2) = 0 The solution is x = 2. Remember to check the solutions you get for log equations. Exercises 15. Solve these equations mentally 11
Example 16. Solve 17. Solve 18. Solve 12
Example 19. 5. Exponential Equations Solving from the definition To solve exponential equations without logarithms, you need to have equations with comparable exponential expressions on either side of the "equals" sign, so you can compare the powers and solve. In other words, you have to have "(some base) to (some power) equals (the same base) to (some other power)", where you set the two powers equal to each other, and solve the resulting equation. For example: Solve 5 x = 5 3. Since the bases ("5" in each case) are the same, then the only way the two expressions could be equal is for the powers also to be the same. That is: x = 3 Solve 10 1 x = 10 4 Since the bases are the same, I can equate the powers and solve: 1 x = 4 1 4 = x 3 = x Sometimes you'll first need to convert one side or the other (or both) to some other base before you can set the powers equal to each other. For example: Solve 3 x = 9. Copyright Elizabeth Stapel 2002-2011 All Rights Reserved 13
Since 9 = 3 2, this is really asking me to solve: 3 x = 3 2 By converting the 9 to a 3 2, I've converted the right-hand side of the equation to having the same base as the left-hand side. Since the bases are now the same, I can set the two powers equal to each other: x = 2 Solving by Using Logarithms Most exponential equations do not solve neatly; in solving these more-complicated equations, you will need to use logarithms. For instance: Solve 2 x = 30. 2 x = 30 ln(2 x ) = ln(30) xln(2) = ln(30) x = ln(30) / ln(2) 14
Exercises 20. Solve by expressing powers with the same base 21. Solve by using logarithms 22. 15
6. Simultaneous Linear Equations A system of two linear equations is just a set of two linear equations. There are a few different methods of solving systems of linear equations: The Substitution Method. First, solve one linear equation for y in terms of x. Then substitute that expression for y in the other linear equation. You'll get an equation in x. Solve this, and you have the x. Then plug in x to either equation to find the corresponding y. (If it's easier, you can start by solving an equation for x in terms of y) Example: Solve the system Solve the second equation for y. y = 19 7x Substitute 19 7x for y in the first equation and solve for x. 3x + 2(19 7x) = 16 3x + 38 14x = 16 11x = 22 x = 2 Substitute 2 for x in y = 19 7x and solve for y. y = 19 7(2) y = 5 The solution is (2, 5). The Linear Combination Method, also known as The Addition Method or The Elimination Method. Add (or subtract) a multiple of one equation to (or from) the other equation, in such a way that either the x-terms or the y-terms cancel out. Then solve for x (or y, whichever's left) and substitute back to get the other coordinate. Example: Solve the system Multiply the first equation by 2 and add the result to the second equation. 8x 6y = 4 8x 2y = 12 8y = 16 Solve for y. y = 2 Substitute for y in either of the original equations and solve for x. 4x + 3( 2) = 2 4x 6 = 2 4x = 4 x = 1 The solution is (1, 2). 16
The Graphing Method. You have just to graph the two lines, and see where they intersect. Example: y = 0.5x + 2 y = 2x 3 (See what we have first expressed y in terms of x in order to graph the lines) The solution is where the two lines intersect, the point ( 2, 1). When plotting the lines on a graph, there are three possibilities: - The lines intersect at zero points. (The lines are parallel and the system has no solutions.) - The lines intersect at exactly one point. (The system has a single solution) - The lines intersect at infinitely many points. (The two equations represent the same line and the system has infinite solutions.) 17
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Exercises 23. Solve 24. Find the number of solutions without solving 25. Solve 26. 19
7. Nonlinear Simultaneous Equations Systems with some nonlinear equations (not of the first degree) Systems with second degree equations They are usually solved by the substitution method Exercises 27. Solve 28. 20
29. Solve. (First square both members and then subtract the equations) 30. 31. 32. 21
Systems of logarithmic and exponential equations They are usually solved by change of variable. 22
Exercises 33. 34. 23
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Key Activities 25
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Exercises 35. Find the number of solutions 36. Solve by factoring Example 37. Solve without using the formula 38. 28
39. Solve 40. Solve 41. 29
42. Solve by using change of variable 43. 44. Solve 45. 46. 30
47. Solve Example 48. Example 31
49. Solve 50. Solve 51. Solve 32
Example 52. Solve 53. 54. 55. Solve 33
56. Solve 57. 58. Solve 59. 34
Example 60. Find the number of solutions without solving 61. 62. 35
Example 63. Solve 64. 65. 36
66. 67. 68. 69. 70. 37
71. 72. 73. 38