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Unformatted text preview: Math 461 F Spring 2011 Homework 2 Solutions Drew Armstrong A.1. Suppose that the cubic equation ax 3 + bx 2 + cx + d = 0 has three roots, called r,s,t . Give a formula for rs + rt + st in terms of a,b,c,d . By the Factor Theorem we can write ax 3 + bx 2 + cx + d = a ( x r )( x s )( x t ) = ax 3 a ( r + s + t ) x 2 + a ( rs + rt + st ) x a ( rst ) . Now recall that two polynomials are equal if and only if their coefficients are equal. Hence we have rs + rt + st = c a . Note that a 6 = 0 because it is the leading coefficient. A.2. Find all complex solutions z ∈ C to the quadratic equation z 2 z + 1 4 i 2 = 0 . Note that z 2 z + 1 4 1 4 + 1 4 i 2 = 0 z 2 z + 1 4 = i 2 z 1 2 2 = i 2 . Hence z 2 1 / 2 must be a square root of i/ 2. There are two of these, and we can find them! Suppose that x 2 = i/ 2 with x = r cis θ in polar form. Thus we have x 2 = r 2 cis(2 θ ) = i 2 = 1 2 cis π 2 . Since the lengths are equal we get r 2 = 1 / 2, or r = 1 / √ 2. Since the angles are equal we get 2 θ = π/ 2 + 2 πk for any integer k . In other words, θ = π/ 4 or θ = 5 π/ 4. We conclude that the square roots of i/ 2 are x = 1 √ 2 cis π 4 = 1 2 + i 2 and x = 1 2 cis 4 π 5 = 1 2 i 2 ....
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This note was uploaded on 01/08/2012 for the course MATH 561 taught by Professor Armstrong during the Spring '11 term at University of Miami.
 Spring '11
 Armstrong
 Algebra, Factor Theorem

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