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Unformatted text preview: Solutions to HW #2 Math 67 UC Davis, Fall 2011 1. (a) We rewrite the second and third equations in the system as (5 z ) x + 2 y = 0 2 x + (8 z ) y = 0 We are looking for a solution other than the zero solution x = y = 0. By a previous exercise mentioned in the problem, such a solution exists if and only if (5 z )(8 z ) 2 2 = 0. This gives the quadratic equation z 2 13 z + 36 = 0, which has the two solutions z = 4 and z = 9. (b) When z = 4, the solution of the original system of three equations is x = 2 ,y = 1. This doesnt make sense since in the original problem x and y referred to the relative importances of two websites, and therefore had to be positive numbers (or 0). When z = 9, the solution is x = 1 / 3 ,y = 2 / 3, which makes sense and is the same solution given in the lecture notes. 2. (a) f ( x ) = x 3 x 2 + 2 , g ( x ) = x 2 = q ( x ) = x 2 + x + 2 ,r ( x ) = 6 (b) f ( x ) = x 3 x 2 + 2 , g ( x ) = x + 1 = q ( x ) = x 2 2 x + 2 ,r ( x ) = 0 (c) f ( x ) = x 4 + x, g ( x ) = x 2 + 1 = q ( x ) = x 2 1 ,r ( x ) = x + 1 3. If p ( x ) = x 3 4 x 2 + 2 x + 3 and p (3) = 0, then p ( x ) is divisible (without remainder)...
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 Fall '07
 Schilling
 Linear Algebra, Algebra, Equations

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