Explanation Formula used: 1) Theorem: Change of variable in the quadratic The change of variable x = y − a 4 in x 4 + a x 3 + b x 2 + c x + d = 0 yields an equation of the form y 4 + p y 2 + q y + r = 0 . The equation t 3 − p t 2 − 4 r t + 4 r p − q 2 = 0 is known as resolvent equation of y 4 + p y 2 + q y + r = 0 . 2) Theorem: Cardano’s Formulas: The solutions to the cubic equation x 3 + a x 2 + b x + c = 0 are given by A 3 + B 3 − a 3 , ω A 3 + ω 2 B 3 − a 3 , and ω 2 A 3 + ω B 3 − a 3 , where ω = − 1 + i 3 2 , p = b − a 2 3 , q = c − a b 3 + 2 a 3 27 , A = − q 2 + ( q 2 ) 2 + ( p 3 ) 3 , B = − q 2 − ( q 2 ) 2 + ( p 3 ) 3 with A 3 and B 3 chosen, so that A 3 B 3 = − p 3 . 3) ( a + b ) 4 = a 4 + 4 a 3 b + 6 a 2 b 2 + 4 a b 3 + b 4 4) ( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 5) ( a + b ) 2 = a 2 + 2 a b + b 2 Explanation: Consider the given equation x 4 + 4 x 3 + 7 x 2 + 8 x + 10 = 0 By using the theorem, the change of variable x = y − a 4 in x 4 + a x 3 + b x 2 + c x + d = 0 yields the equation y 4 + p y 2 + q y + r = 0 By comparing x 4 + 4 x 3 + 7 x 2 + 8 x + 10 = 0 with x 4 + a x 3 + b x 2 + c x + d = 0 gives a = 4 . Therefore, substitute x = y − 4 4 = y − 1 in x 4 + 4 x 3 + 7 x 2 + 8 x + 10 = 0 as a = 4 . ( y − 1 ) 4 + 4 ( y − 1 ) 3 + 7 ( y − 1 ) 2 + 8 ( y − 1 ) + 10 = 0 Simplify by using formulas, ⇒ y 4 − 4 y 3 + 6 y 2 − 4 y + 1 + 4 ( y 3 − 3 y 2 + 3 y − 1 ) + 7 ( y 2 − 2 y + 1 ) + 8 y − 8 + 10 = 0 Combine like terms, ⇒ y 4 + ( − 4 + 4 ) y 3 + ( 6 − 12 + 7 ) y 2 + ( − 4 + 12 − 14 + 8 ) y + ( 1 − 4 + 7 − 8 + 10 ) = 0 ⇒ y 4 + ( 0 ) y 3 + ( 1 ) y 2 + ( 2 ) y + ( 6 ) = 0 ⇒ y 4 + y 2 + 2 y + 6 = 0 Here, p = 1 , q = 2 and r = 6 The resolvent equation is given by, t 3 − p t 2 − 4 r t + 4 r p − q 2 = 0 t 3 − ( 1 ) t 2 − 4 ( 6 ) t + 4 ( 6 ) ( 1 ) − ( 2 ) 2 = 0 ⇒ t 3 − t 2 − 24 t + 24 − 4 = 0 ⇒ t 3 − t 2 − 24 t + 20 = 0 Solve this by using Cardano’s Formulas, Here, a = − 1 , b = − 24 , and c = − 20

8th Edition

ISBN: 9781285463230

Author: Gilbert, Linda, Jimmie

Publisher: Cengage Learning,

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Chapter 8.5, Problem 22E

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The solution of the equation x4+ 4x3+7x2+ 8x+10=0.

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Chapter 8 Solutions

Elements Of Modern Algebra

Ch. 8.1 - True or False Label each of the following...Ch. 8.1 - Prob. 2TFE Ch. 8.1 - Prob. 3TFE Ch. 8.1 - Prob. 4TFE Ch. 8.1 - Prob. 5TFE Ch. 8.1 - Prob. 6TFE Ch. 8.1 - Prob. 7TFE Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E

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The solution of the equation x 4 + 4 x 3 + 7 x 2 + 8 x + 10 = 0 .

Solution Summary: The author explains how the solution of the cubic equation x3+a2 +bx+c=0 can be determined.

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The solution of the equation x4+ 4x3+7x2+ 8x+10=0.

Chapter 8 Solutions