1873_solutions

# Boil down to the one possibility that r 3 3 and s 3 9

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Unformatted text preview: rvation leads to x 1 x2  x 2  0 giving us the solutions x  1 or x  1 or x  2. On the other hand, if we look for r and s to make r3  s3  2 3rs  3 then we obtain r3  3 1 r 2 which becomes r 6 2r 3  1  0 which leads to the equation r  1. Thus r  s  1 and we obtain the solutions x r s 2 or rs 3i r s  2 0  1. x 2 2 d. x 3 6x  10  0 We look for numbers r and s such that r 3  s 3  10 3rs  6 which gives us r3  3 2 r  10 in other words, 10r 3  8  0 r6 and we obtain r3  5 and we may assume that r  3 5  17 and s  x r s 3 17 17 3 17 . Thus we have the possibilities 5 5 3 17  5 or x rs 3i r 2 s 3  5  17  3 5 17 3i 3 5  17 3 5 17 2 e. x 3 6x 2  6x  14  0. We begin by making the substitution u  x x 3 6x 2  6x  14  0 becomes 15 c and the equation . uc 6 uc 3 2  6 u  c  14  0 giving us u 3  3c 6 u 2  3c 2 12c  6 u  c 3 6c 2  6c  14  0 and now we see that the definition c  2 will convert the equation into a form with no quadratic term. We obtain u 3 6u  10  0 We now copy the solution obtained in part d and obtain u 17 3 5 17  5 3 or 3 u 5  17  3 5 17 3 5  17 3 3i 17  5 3 5 17 2 and so x  2 17 3 5 or 3 x  2 5  17  3 5 17 3i 3 5  17 2 2. Show that if w 3  k 2 then the cubic equation 3wx  2k  0 x3 has one real solution x 3 k2 w3 w3  k k2 3 k We look for numbers r and s such that r 3  s 3  2k 3rs  3w which gives us r 3 k r 3 k  k2 w3 s 3 k k2 w3 k2 w3 . In other words, we can assume that and and we conclude that the only real solution of the equation x 3 3wx  2k  0 is x 3 k2 w3 k 3 k2 w3  k . 3. By considering the solution of the equation x 1 x2  x  3  0 (or otherwise) deduce that 3 9 3  5 11  63 16 93 3 5 11 63  1. 3 5 17 . We see at once that the only real solution of the equation x 1 x2  x  3  0 is x  1. On the other hand, the equation can be written as x 3  2x 3  0 and we can solve by looking for numbers r and s satisfying r3  s3  3 3rs  2 From which we obtain r3  3 2 3r 3 which gives us r3  9 3 5 11 63 r 3 9 3  5 11 63 r 3 9 3 5 11 63 from which we can select and which gives us the real solution of the equation x 1 x2  x  3  0 as x r s 3 9 3  5 11  63 93 3 5 11 63 and we conclude that 3 4. 9 3  5 11  63 93 3 5 11 63  1. a. Show that if  and  are given numbers and if the equation x  x 2  x    0 is written in the form x 3  ax  b  0 then a   Since 2 and b  . x  x 2  x    x 3   2 x  the desired result follows at once. b. Given that  and  are real numbers and that  2  4, explain why the equation x  x 2  x    0 has only one real solution. By considering this equation, or otherwise, deduce that 3 3   2 2   4 2 3 63 With a    2 and b   the inequality 17  3 3  2 2   3 63 4 2  . 27b 2  4a 3  0 says that which holds if and only if such that 2 2  2   2 2 4  4. Now suppose that 0 2  4. We look for two numbers r and s r 3  s 3   3rs   2 giving us 3 2  3r r3    27r 6 ...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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