Contents 16. Higher Degree Equations

Transcription

1 Contents 16. Higher Degree Equations Finding Roots of Higher Degree Equations Example Example Example Rational Functions Example

2 Peterson, Technical Mathematics, 3rd edition Finding Roots of Higher Degree Equations In Section 16.2, we determined the number of roots of a polynomial. We also learned that complex roots come in pairs and that once we are able to reduce the nonlinear factor to degree 2, we can use the quadratic formula. The TI-83 graphing calculator allows you to solve any polynomial of degrees This is done by installing the program the PolySmlt 1. To use this program, which TI refers to as an application, press apps and use the to scroll down until the cursor is on PolySmlt and press ENTER. You will see a screen that shows the Texas Instruments logo, the name and version of the program. Press any key or just wait and you will get the MAIN MENU screen with several options. Select option 1:PolyRootFinder. You will be presented with the line Degree of Poly = and you should enter the degree of the polynomial. You will then be asked to enter the coefficients starting with a n, the coefficient of the x n term, through a 0, the constant term. Then press the GRAPH which selects the SOLVE option at the bottom of the screen and after a pause the n roots will be displayed. To solve another polynomial press the Y= to select the MAIN option. Example Find all the roots of P (x) = 3x 3 x 2 10x + 8. Solution In this polynomial, P (x) = 3x 3 x 2 10x + 8, n = 3, so the degree of the polynomial is 3. We also have a 3 = 3, a 2 = 1, a 1 = 10, and a 0 = 8. The data is entered as shown in Figure 16.1a and the solution looks like that in Figure 16.1b. From Figure 16.1b we see that the roots are x 1 = 1, x 2 = , and x 3 = 2. If you want to save any of these solutions put the cursor on the line of the one you want to save, press STO and then the letter of the variable where you want the value stored and press ENTER. Two of the solutions seem very clear. We suspect that x 2 = is probably x 2 = = 4 3. To see if this is correct we store 4 3 as x and then key in 3x 3 x 2 10x + 8. When we press enter the result is 0 and this confirms that x 2 = = 4 3. Thus, using the factor theorem, we can write P (x) as P (x) = 3(x + 2) ( x 3) 4 (x 1). FIGURE 16.1a FIGURE 16.1b Example Find all the roots of P (x) = 4x 3 5x 2 2x To download this software you need the going to the webpage. You must use TI TM Connect or TI-GRAPH LINK TM software and a TI-GRAPH LINK cable to install the application.

3 Peterson, Technical Mathematics, 3rd edition 3 Solution In this polynomial, P (x) = 4x 3 5x 2 2x + 3 with n = 3, so the order of the polynomial is 3. We also have a 3 = 4, a 2 = 5, a 1 = 2, and a 0 = 3. The data is entered as in Example and the solution looks like that in Figure 16.2a. Here the solutions are the real number x 1 =.75 and the nonreal complex numbers x 2 = E 7i and x 3 = E 7i. FIGURE 16.2a We know that x 1 is the real number 0.75 = 3 4. How do we know that x 2 and x 3 are not real numbers? According to the calculator, x 2 = j = j. This is very close to 1. If we look at the graph of this polynomial in the neighborhood of these roots we get Figure 16.2b. We can see that x 1 = 0.75 is one place where the graph crosses the x-axis and it seems as if the graph just touches, but does not cross, the x-axis at x = 1. However, the resolution of the graphing calculator makes it difficult to tell. We need to check to see whether x 2 = 1 or x 2 = j = j is the answer. [ 1, 1.5] [ 2, 4] We will begin by checking x = 1. Store 1 as x by pressing 1 STO FIGURE 16.2b x-var ENTER. Now, key in 4x 3 5x 2 2x + 3 ENTER. The result is 0 which means that P (1) = 0. So, x 1 = 1 is a solution. Similarly, we can verify that x 2 = 1 and 1 is a double root and this polynomial has three real roots. In factored form P (x) = (x )(x 1) 2 = (4x + 3)(x 1) 2. Example Determine all the roots of P (x) = 5x 6 4x 5 41x x x 2 28x 7. Solution In this polynomial n = 6. Entering the coefficients a 6 = 5, a 5 = 4, a 4 = 41,..., a 1 = 28, and a 0 = 7 into the calculator we get the following solutions x 1 = 0.2 x 2 = 1 x 3 = x 4 = x 5 = E 7i x 6 = E 7i All the roots seem to be between 2.7 and 2.7. If we graph P on this interval and use ZFIT we obtain the graph in Figure Here we can clearly see that there are real roots at x 1 = 0.2, x 2 = 1, x , and x It appears that x 5 = x 6 = 1 is a double root and if we check we see that these are the roots. Thus, P (x) 5(x )(x )(x + 1)(x 1) 2 (x + 0.2) FIGURE 16.3 In most technical areas the decimal approximations of x 3 and x 4 given by the calculator are sufficient. If you needed to determine the exact values of the roots x and x of the polynomial P (x) = 5x 6 4x 5 41x x x 2 28x 7 you could use synthetic division using the first four roots and then use the quadratic formula on the resulting quadratic factor.

4 Peterson, Technical Mathematics, 3rd edition 4 Thus, we see that P (x) = (x + 1)(x 1) 2 (x + 0.2)(5x 2 35) = 5(x + 1)(x 1) 2 (x + 0.2)(x 2 7) Since the solution to x 2 7 = 0 is x = ± 7 we see that the exact values of x 1 and x 2 in Example are x 1 = 7 and x 2 = 7.

5 Peterson, Technical Mathematics, 3rd edition Rational Functions Example Use a graphing calculator to solve 6 x + 1 = 4 x + 2. Solution If you begin by graphing y1 = 6 4 and y2 = on the same set of x + 1 x + 2 axes you get a result something like the one in Figure 16.14a. As you can see, this is a mess: it is very difficult to read and will not be of much use in finding a solution. FIGURE 16.14a FIGURE 16.14b Graphing y1 y2 on the standard viewing window produces Figure 16.14b. This is still a difficult figure to read, but it has one major advantage over Figure 16.14a. In Figure 16.14a we wanted to find where the graphs of y1 = 6 and y2 = 4 x + 2 x + 1 intersected. In Figure 16.14b we are looking for the places where y1 y2 crosses the x-axis. If we zoom in a little by just changing the y-values of the viewing window to ymin= 2 and ymax= 2 we get Figure 16.14c. It looks as if this graph crosses the x-axis three times, near x = 4, x = 2, and x = 1. But, the latter two, x = 2 and x = 1, are vertical asymptotes and cannot be solutions. FIGURE 16.14c FIGURE 16.14d Here the PolySmlt application of the calculator will not help. We need to resort to the zero feature we first explored in Section 4.5. Press 2nd TRACE [CALC] 2 [2: zero]. After selecting the left bound, right bound, and a guess we see, in Figure 16.14d, that the solution is x = 4.