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2 48 Chapter 5 Quadratic Functions x = ±5 These solutions are called square roots of 5. Because there are two solutions, we need a different notation for each. We will denote the positive square root of 5 with the notation 5 and the negative square root of 5 with the notation 5. Thus, 5 = 5 and 5 = 5. In a similar vein, the equation x = 36 has two solutions, x = ± 36, or alternatively, x = ±6. The notation 36 calls for the positive square root, while the notation 36 calls for the negative square root. That is, 36 = 6 and 36 = 6. It is not necessary that the right-hand side of the equation x = a be a perfect square. For example, the equation x = 7 has solutions x = ± 7. (4) There is no rational square root of 7. That is, there is no way to express the square root of 7 in the form p/q, where p and q are integers. Therefore, 7 is an example of an irrational number. However, 7 is a perfectly valid real number and we re perfectly comfortable leaving our answer in the form shown in equation (4). However, if an approximation is needed for the square root of 7, we can reason that because 7 lies between 4 and 9, the square root of 7 will lie between and 3. Because 7 is closer to 9 than 4, a reasonable approximation might be A calculator can provide an even better approximation. For example, our TI83 reports There are two degenerate cases involving the equation x = a that demand our attention. 1. The equation x = 0 has only one solution, namely x = 0. Thus, 0 = 0.. The equation x = 4 has no real solutions. 4 It is not possible to square a real number and get 4. In this situation, we will simply state that the equation x = 4 has no real solutions (no solutions that are real numbers). 3 The symbol means approximately equal to. 4 It is incorrect to state that the equation x = 4 has no solutions. If we introduce the set of complex numbers (a set of numbers introduced in college algebra and trigonometry), then the equation x = 4 has two solutions, both of which are complex numbers. Version: Fall 007

3 Section 5.4 The Quadratic Formula 483 Example 5. Find all real solutions of the equations x = 30, x = 0, and x = 14. The solutions follow. The equation x = 30 has two real solutions, namely x = ± 30. The equation x = 0 has one real solution, namely x = 0. The equation x = 14 has no real solutions. Let s try additional examples. Example 6. Find all real solutions of the equation (x + ) = 43. There are two possibilities for x +, namely x + = ± 43. To solve for x, subtract from both sides of this last equation. x = ± 43. Although this last answer is usually the preferable form of the answer, there are some times when an approximation is needed. So, our TI83 gives the following approximations and Example 7. Find all real solutions of the equation (x 4) = 15. If x is a real number, then so is x 4. It s not possible to square the real number x 4 and get 15. Thus, this problem has no real solutions. 5 Development of the Quadratic Formula We now have all the groundwork in place to pursue a solution of the general quadratic equation ax + bx + c = 0. (8) We re going to use a form of completing the square to solve this equation for x. Let s begin by subtracting c from both sides of the equation. ax + bx = c 5 Again, when you study the complex numbers, you will learn that this equation has two complex solutions. Hence, it is important that you do not say that this problem has no solutions, as that is simply not true. You must say that this problem has no real solutions. Version: Fall 007

4 484 Chapter 5 Quadratic Functions Next, divide both sides of the equation by a. x + b a x = c a Take half of the coefficient of x, as in (1/)(b/a) = b/(a). Square this result to get b /(4a ). Add this amount to both sides of the equation. x + b a x + b 4a = c a + b 4a On the left we factor the perfect square trinomial. On the right we get a common denominator and add the resulting equivalent fractions. ( x + b ) = 4ac a 4a + b 4a ( x + b ) = b 4ac a 4a Provided the right-hand side of this last equation is positive, we have two real solutions. x + b b a = ± 4ac 4a On the right, we take the square root of the top and the bottom of the fraction. 6 x + b b a = ± 4ac a To complete the solution, we need only subtract b/(a) from both sides of the equation. x = b b a ± 4ac a Although this last answer is a perfectly good solution, we customarily rewrite the solution with a single common denominator. x = b ± b 4ac a This last result gives the solution to the general quadratic equation (8). The solution (9) is called the quadratic formula. (9) 6 In a later section we will present a more formal approach to the symbolic manipulation of radicals. For now, you can compute (/3) with the calculation (/3)(/3) = 4/9, or you can simply square numerator and denominator of the fraction, as in (/3) = ( /3 ) = 4/9. Conversely, one can take the square root of a fraction by taking the square root of the numerator divided by the square root of the denominator, as in 4/9 = 4/ 9 = /3. Version: Fall 007

5 Section 5.4 The Quadratic Formula 485 The Quadratic Formula. The solutions to the quadratic equation are given by the quadratic formula ax + bx + c = 0 (10) x = b ± b 4ac. (11) a Although the development of the quadratic formula can be intimidating, in practice its application is quite simple. Let s look at some examples. Example 1. Use the quadratic formula to solve the equation x = 7 6x. The first step is to place the equation in the form ax + bx + c = 0 by moving every term to one side of the equation, 7 arranging the terms in descending powers of x. x + 6x 7 = 0 Next, compare x + 6x 7 = 0 with the general form of the quadratic equation ax + bx + c = 0 and note that a = 1, b = 6, and c = 7. Copy down the quadratic formula. x = b ± b 4ac a Substitute a = 1, b = 6, and c = 7 and simplify. x = (6) ± (6) 4(1)( 7) (1) x = 6 ± x = 6 ± 144 In this case, 144 is a perfect square. That is, 144 = 1, so we can continue to simplify. x = 6 ± 1 It s important to note that there are two real answers, namely Simplifying, x = 6 1 or x = x = 9 or x = We like to say Make one side equal to zero. Version: Fall 007

7 Section 5.4 The Quadratic Formula 487 Simplify. x = ± x = ± 1 In this case, 1 is not a perfect square, so we ve simplified as much as is possible at this time. 10 However, we can approximate these solutions with the aid of a calculator. x = and x = + 1 We will find these approximations useful in what follows (14) The equations in Examples 1 and 13 represent a fundamental shift in our usual technique for solving equations. In the past, we ve tried to isolate the terms containing x (or whatever unknown we are solving for) on one side of the equation, and all other terms on the other side of the equation. Now, in Examples 1 and 13, we find ourselves moving everything to one side of the equation, making one side of the equation equal to zero. This bears some explanation. Linear or Nonlinear. Let s assume that the unknown we are solving for is x. If the highest power of x present in the equation is x to the first power, then the equation is linear. Thus, for example, each of the equations x + 3 = 7, 3 4x = 5x + 9, and ax + b = cx + d is linear. If there are powers of x higher than x to the first power in the equation, then the equation is nonlinear. Thus, for example, each of the equations is nonlinear. x 4x = 9, x 3 = x + 3, and ax + bx = cx + d The strategy for solving an equation will shift, depending on whether the equation is linear or nonlinear. 10 In a later chapter on irrational functions, we will take up the topic of simplifying radical expressions. Until then, this form of the final answer will have to suffice. Version: Fall 007

8 488 Chapter 5 Quadratic Functions Solution Strategy Linear Versus Nonlinear. When solving equations, you must first ask if the equation is linear or nonlinear. Again, let s assume the unknown we wish to solve for is x. If the equation is linear, move all terms containing x to one side of the equation, all the remaining terms to the other side of the equation. If the equation is nonlinear, move all terms to one side of the equation, making the other side of the equation zero. Thus, because ax + b = cx + d is linear in x, the first step in solving the equation would be to move all terms containing x to one side of the equation, all other terms to the other side of the equation, as in ax cx = d b. On the other hand, the equation ax + bx = cx + d is nonlinear in x, so the first step would be to move all terms to one side of the equation, making the other side of the equation equal to zero, as in ax + bx cx d = 0. In Example 13, the equation x x = is nonlinear in x, so we moved everything to the left-hand side of the equation, making the right-hand side of the equation equal to zero, as in x x = 0. However, it doesn t matter which side you make equal to zero. Suppose instead that you move every term to the right-hand side of the equation, as in 0 = x + x +. Comparing 0 = x + x + with general quadratic equation 0 = ax + bx + c, note that a = 1, b =, and c =. Write down the quadratic formula. x = b ± b 4ac a Next, substitute a = 1, b =, and c =. Again, note the careful use of parentheses. This leads to two solutions, x = () ± () 4( 1)() ( 1) x = ± = ± 1. In Example 13, we found the following solutions and their approximations. x = and x = Version: Fall 007

10 490 Chapter 5 Quadratic Functions It will now be apparent why we used our calculator to approximate the solutions in (14). These are the x-coordinates of the x-intercepts. One x-intercept is located at approximately ( 0.73, 0), the other at approximately (.73, 0). These approximations are used to plot the location of the intercepts as shown in Figure 1(b). However, the actual values of the intercepts are (( 1)/, 0) and ((+ 1)/, 0), and these exact values should be used to annotate the intercepts, as shown in Figure 1(b). Finally, to find the y-intercept, let x = 0 in g(x) = x x. Thus, g(0) = and the y-intercept is (0, ). The y-intercept and its mirror image across the axis of symmetry are both plotted in Figure 1(c), where the final graph of the parabola is also shown. y y y (1, 3) x ( ) ( ) 1, 0 + 1, 0 x (1, 3) ( ) ( ) 1, 0 + 1, 0 x (0, ) (1, 3) x = 1 (a) Plotting the vertex and axis of symmetry. x = 1 (b) Adding the x-intercepts provides added accuracy. x = 1 (c) Adding the y-intercept and its mirror image provides an excellent final graph. Figure 1. We ve made an important point and we pause to provide emphasis. Zeros and Intercepts. Whenever you use the quadratic formula to solve the quadratic equation the solutions are the zeros of the quadratic function ax + bx + c = 0, x = b ± b 4ac a f(x) = ax + bx + c. The solutions also provide the x-coordinates of the x-intercepts of the graph of f. We need to discuss one final concept. Version: Fall 007

11 Section 5.4 The Quadratic Formula 491 The Discriminant Consider again the quadratic equation ax + bx + c = 0 and the solutions (zeros) provided by the quadratic formula x = b ± b 4ac. a The expression under the radical, b 4ac, is called the discriminant, which we denote by the letter D. That is, the formula for the discriminant is given by D = b 4ac. The discriminant is used to determine the nature and number of solutions to the quadratic equation ax +bx+c = 0. This is done without actually calculating the solutions. Let s look at three key examples. Example 15. Consider the quadratic equation x 4x 4 = 0. Calculate the discriminant and use it to determine the nature and number of the solutions. Compare x 4x 4 = 0 with ax + bx + c = 0 and note that a = 1, b = 4, and c = 4. The discriminant is given by the calculation D = b 4ac = ( 4) 4(1)( 4) = 3. Note that the discriminant D is positive; i.e., D > 0. Consider the quadratic function f(x) = x 4x 4, which can be written in vertex form f(x) = (x ) 8. This is a parabola that opens upward. It is shifted to the right units, then downward 8 units. Therefore, it will cross the x-axis in two locations. Hence, one would expect that the quadratic formula would provide two real solutions (x-intercepts). Indeed, x = ( 4) ± ( 4) 4(1)( 4) (1) = 4 ± 3. Note that the discriminant, D = 3 as calculated above, is the number under the square root. These solutions have approximations x = and x = , which aid in plotting an accurate graph of f(x) = (x ) 8, as shown in Figure. Version: Fall 007

12 49 Chapter 5 Quadratic Functions y ( ) ( ) 4 3, , 0 x (, 8) Figure. If the discriminant is positive, there are two real x-intercepts. Thus, if the discriminant is positive, the parabola will have two real x-intercepts. Next, let s look at an example where the discriminant equals zero. Example 16. Consider again the quadratic equation ax + bx + c = 0 and the solutions (zeros) provided by the quadratic formula x = b ± b 4ac. a The expression under the radical, b 4ac, is called the discriminant, which we denote by the letter D. That is, the formula for the discriminant is given by D = b 4ac. The discriminant is used to determine the nature and number of solutions to the quadratic equation ax +bx+c = 0. This is done without actually calculating the solutions. Consider the quadratic equation x 4x + 4 = 0. Calculate the discriminant and use it to determine the nature and number of the solutions. Compare x 4x + 4 = 0 with ax + bx + c = 0 and note that a = 1, b = 4, and c = 4. The discriminant is given by the calculation Note that the discriminant equals zero. D = b 4ac = ( 4) 4(1)(4) = 0. Consider the quadratic function f(x) = x 4x + 4, which can be written in vertex form Version: Fall 007

13 Section 5.4 The Quadratic Formula 493 f(x) = (x ). (17) This is a parabola that opens upward and is shifted units to the right. Note that there is no vertical shift, so the vertex of the parabola will rest on the x-axis, as shown in Figure 3. In this case, we found it necessary to plot two points to the right of the axis of symmetry, then mirror them across the axis of symmetry, in order to get an accurate plot of the parabola. y 10 (, 0) x 10 x f(x) = (x ) x = Figure 3. At the right is a table of points satisfying f(x) = (x ). These points and their mirror images are seen as solid dots superimposed on the graph of f(x) = (x ) at the left. Take a closer look at equation (17). If we set f(x) = 0 in this equation, then we get 0 = (x ). This could be written 0 = (x )(x ) and we could say that the solutions are and again. However, mathematicians prefer to say that is a solution of multiplicity or is a double solution. 11 Note how the parabola is tangent to the x-axis at the location of the double solution. That is, the parabola comes down from positive infinity, touches (but does not cross) the x-axis at x =, then rises again to positive infinity. Of course, the situation would be reversed in the parabola opened downward, as in g(x) = (x ), but the graph would still kiss the x-axis at the location of the double solution. Still, the key thing to note here is the fact that the discriminant D = 0 and the parabola has only one x-intercept. That is, the equation x 4x + 4 = 0 has a single real solution. Next, let s look what happens when the discriminant is negative. Example 18. Consider the quadratic equation x 4x + 8 = Actually, mathematicians call these double roots, but we prefer to postpone that language until the chapter on polynomial functions. Version: Fall 007

14 494 Chapter 5 Quadratic Functions Calculate the discriminant and use it to determine the nature and number of the solutions. Compare x 4x + 8 = 0 with ax + bx + c = 0 and note that a = 1, b = 4, and c = 8. The discriminant is given by the calculation Note that the discriminant is negative. D = b 4ac = ( 4) 4(1)(8) = 16. Consider the quadratic function f(x) = x 4x + 8, which can be written in vertex form f(x) = (x ) + 4. This is a parabola that opens upward. Moreover, it has to be shifted units to the right and 4 units upward, so there can be no x-intercepts, as shown in Figure 4. Again, we found it necessary in this example to plot two points to the right of the axis of symmetry, then mirror them, in order to get an accurate plot of the parabola. y 10 (, 4) x f(x) = (x ) +4 x x = Figure 4. At the right is a table of points satisfying f(x) = (x ) + 4. These points and their mirror images are seen as solid dots superimposed on the graph of f(x) = (x ) + 4 at the left. Once again, the key point in this example is the fact that the discriminant is negative and there are no real solutions of the quadratic equation (equivalently, there are no x-intercepts). Let s see what happens if we actually try to find the solutions of x 4x + 8 = 0 using the quadratic formula. Again, a = 1, b = 4, and c = 8, so x = b ± b 4ac a = ( 4) ± ( 4) 4(1)(8). (1) Simplifying, x = 4 ± 16. Version: Fall 007

15 Section 5.4 The Quadratic Formula 495 Again, remember that the number under the square root is the discriminant. In this case the disriminant is 16. It is not possible to square a real number and get 16. Thus, the quadratic equation x 4x + 8 = 0 has no real solutions, as predicted. Let s summarize the findings in our last three examples. Summary 19. Consider the quadratic equation ax + bx + c = 0. The discriminant is defined as D = b 4ac. There are three possibilities: 1. If D > 0, then the quadratic equation has two real solutions.. If D = 0, then the quadratic equation has one real solution. 3. If D < 0, then the quadratic equation has no real solutions. This key result is reflected in the graph of the quadratic function. Summary 0. Consider the quadratic function f(x) = ax + bx + c. The graph of this function is a parabola. Three possibilities exist depending upon the value of the discriminant D = b 4ac. 1. If D > 0, the parabola has two x-intercepts.. If D = 0, the parabola has exactly one x-intercept. 3. If D < 0, the parabola has no x-intercepts. Version: Fall 007

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