1873_solutions

# Given that and are real numbers and that 2 4 explain

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Unformatted text preview: 27r 3   2  3 0 and we can take 4 r 3 s 3 2 4 3 3     2 2 2 6 3  and 3 3    2 2 6 3  then the real solution of the equation  x 2  x    0 x appears as r s 3 3   2 2   4 2 3 63  3 3  2 2   4 2 3 63 . Since that real solution has to be  we obtain 3 3   2 2   4 2 3 63  3 3  2 2   4 3 63 2  . c. Given that  and  are real numbers and that  2  4, consider the two complex roots of the equation x  x 2  x    0 and deduce that 3 3   2 2   4 2 3 63 1 2 2 4  2 2 4 3 and 3 3  2 2   4 2 3 63 The complex roots are rs 3i r 2 s and since these two complex roots are also  18 i 4 2 2 1 2 2 3 we deduce that rs   and 4 s r By looking at the values of r and s we see that r actually 3 s  0 and so the preceding equation is 4 s r 2 2 3 . Adding we obtain 4 2r    2 3 which gives us 3 3  2 2   4 2 3 63 1 2 4 1 2 2 4  2 2 3 and by subtracting we obtain 3 3   2 2   4 2 3 63 2 3 Some Exercises on the Trigonometric Method 1. By considering the equation x 2 x3  0 1x show that x3 27 cos 3 7x  6  0 1 arccos 3 93 77 2 and 27 cos 3 1 arccos 3 93 77  2 3 3 27 cos 3 1 arccos 3 93 77 2 3 1 and Solution: We observe first that x 1x 2 x  3  x3 7x  6. To solve the equation x 3 7x  6  0 we look for a solution of the form x  k cos  and express the equation as k 3 cos 3  7k cos   6 and we choose k in such a way that 19 k3  4 3 7k which will hold when 27 . 3 k The equation 7x  6  0 x3 now becomes 3 27 3 cos 3  27 3 7 cos   6 which gives us 4 cos 3  93 77 3 cos   in other words, 93 77 cos 3  and we deduce that either x 27 cos 3 93 77 1 arccos 3 or x 27 cos 3 1 arccos 3 93 77  2 3 x 27 cos 3 1 arccos 3 93 77 2 3 or Finally we observe that since 0  1 arccos 3 1 arccos 3  3 93 77 93 77 we have cos 1 2 and cos 93 77 1 arccos 3  2 3 0 2 3 1 2 and cos 93 77 1 arccos 3 Thus the largest of the solutions of the equation is 27 cos 3 1 arccos 3 93 77 and the negative solution is 27 cos 3 1 arccos 3 20 93 77  2 3 and the remaining solution is 27 cos 3 1 arccos 3 93 77 2 3 2. Given real numbers , and , prove that the number 2    2   2 3 1 arccos 3 cos is the largest of the three numbers  and  and 3 3  2    2 2    2   2    2   2  . Hint: Apply the method of Exercise 1 to the equation x x  x      0. Exercises on Proof by Contradiction 1. Prove that the following numbers are irrational: a. log 10 5 To obtain a contradiction, assume that log 10 5 is rational and choose positive integers m and n such that log 10 5  m n Since 10 m/n  5 we have 10 m  5 n which is impossible because 10 m is even and 5 n is odd. b. log 12 24 Solution: To obtain a contradiction, assume that the number log 12 24 is rational and choose positive integers m and n such that log 12 24  m . n Since 12 m/n  24 we have 12 m  24 n which we can express as 12 m n  2 n Since 2 n  1 we see that m  n and so the...
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