College Algebra Exam Review 309

College Algebra - .v 3 ± u 3 C q D 3uv D p Now using the second equation to eliminate u from the first gives a qua-dratic equation for v 3 v 3 ±

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7.2. SOLVING THE CUBIC EQUATION 319 It is simpler to deal with a polynomial with zero quadratic term, and this can be accomplished by a linear change of variables y D x C a=3 . We compute that the equation is transformed into y 3 C . ± a 2 3 C b/ y C . 2a 3 27 ± ab 3 C c/: Changing notation, we can suppose without loss of generality that we have at the outset a polynomial equation without quadratic term f.x/ D x 3 C px C q D 0; with roots ˛ 1 2 3 satisfying ˛ 1 C ˛ 3 C ˛ 3 D 0 ˛ 1 ˛ 2 C ˛ 1 ˛ 3 C ˛ 2 ˛ 3 D p ˛ 1 ˛ 2 ˛ 3 D ± q: (7.2.1) If we experiment with changes of variables in the hope of somehow simplifying the equation, we might eventually come upon the idea of ex- pressing the variable x as a difference of two variables x D v ± u . The result is .v 3 ± u 3 / C q ± 3uv.v ± u/ C p.v ± u/ D 0: A sufficient condition for a solution is
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Unformatted text preview: .v 3 ± u 3 / C q D 3uv D p: Now, using the second equation to eliminate u from the first gives a qua-dratic equation for v 3 : v 3 ± p 3 27v 3 C q D 0: The solutions to this are v 3 D ± q 2 ˙ s q 2 4 C p 3 27 : It turns out that we get the same solutions to our original equation regardless of the choice of the square root. Let ! denote the primitive third root of unity ! D e 2±i=3 , and let A denote one cube root of ± q 2 C s q 2 4 C p 3 27 : Then the solutions for v are A; !A , and ! 2 A , and the solutions for x D v ± p 3v are ˛ 1 D A ± p 3A ; a 2 D !A ± ! 2 p 3A ; ˛ 3 D ! 2 A ± ! p 3A :...
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.

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