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ProblemSet2 Solutions

# ProblemSet2 Solutions - Department of Economics Ryerson...

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Department of Economics Ryerson University ECN230-Mathematics For Economics Problem Set #2 - Solutions EXERCISE 3.3 Problem 3: (a) Cubic Equation: ( x x 1 ) ( x x 2 ) ( x x 3 ) = 0 ( x 6) ( x + 1) ( x 3) = 0 ¡ x 2 5 x 6 ¢ ( x 3) = 0 x 3 8 x 2 + 9 x + 18 = 0. (b) Quartic Equation: ( x x 1 ) ( x x 2 ) ( x x 3 ) ( x x 4 ) = 0 ( x 1) ( x 2) ( x 3) ( x 5) = 0 ¡ x 2 3 x + 2 ¢ ( x 3) ( x 5) = 0 ¡ x 3 6 x 2 + 11 x 6 ¢ ( x 5) = 0 x 4 11 x 3 + 41 x 2 61 x + 30 = 0. Problem 4: (a) x = 1 is a root: 1 + ( 2) + 3 + ( 2) = 0. (b) x = 1 is not a root: 2 + ( 1 2 ) + 1 + ( 2) = 1 2 6 = 0. (c) x = 1 is a root: 3 + ( 1) + 2 + ( 4) = 0. Problem 5: (a) Divisors of a 0 = 6 : r = { 1 , 1 , 2 , 2 , 3 , 3 , 6 , 6 } Divisors of a 3 = 1 : s = { 1 , 1 } 1

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Possible rational roots: r s = { 1 , 1 , 2 , 2 , 3 , 3 , 6 , 6 } Roots: x 1 = 1 , x 2 = 2 , x 3 = 3 (b) Divisors of a 0 = 1 : r = { 1 , 1 } Divisors of a 3 = 8 : s = { 1 , 1 , 2 , 2 , 4 , 4 , 8 , 8 } Possible rational roots: r s = © 1 , 1 , 1 2 , 1 2 , 1 4 , 1 4 , 1 8 , 1 8 ª Roots: x 1 = 1 , x 2 = 1 2 , x 3 = 1 4 (c) Multiplying both sides by 8 yields: 8 x 3 + 6 x 2 3 x 1 = 0. The resulting equation is the same as the one in part (b) above. (d) [Note: There is a typo in the textbook regarding the sign of the coe cient of x . It should be + 3 2 instead of 3 2 .] Multiplying both sides by 4 yields: 4 x 4 24 x 3 + 31 x 2 + 6 x 8 = 0.
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