Explanation Formula Used: 1. a 2 − b 2 = ( a − b ) ( a + b ) 2. ( a + b ) 2 = a 2 + 2 a b + b 2 3. Quadratic formula: Consider a quadratic equation a x 2 + b x + c = 0 , its solutions are given by: x = − b ± b 2 − 4 a c 2 a Calculations: Consider the quartic equation x 4 − 4 x = 1 . Since, it is a polynomial of degree 4, so it has 4 roots. Now add 2 x 2 + 1 to both sides of this equation. x 4 − 4 x + 2 x 2 + 1 = 1 + 2 x 2 + 1 x 4 + 2 x 2 + 1 = 2 x 2 + 4 x + 2 x 4 + 2 x 2 + 1 2 = 2 ( x 2 + 2 x + 1 ) ( x 2 + 1 ) 2 = 2 ( x + 1 ) 2 [ ∵ ( a + b ) 2 = a 2 + 2 a b + b 2 ] ( x 2 + 1 ) 2 = [ 2 ( x + 1 ) ] 2 ( x 2 + 1 ) 2 − [ 2 ( x + 1 ) ] 2 = 0 Now, apply the identity a 2 − b 2 = ( a − b ) ( a + b ) [ x 2 + 1 + 2 ( x + 1 ) ] [ x 2 + 1 − 2 ( x + 1 ) ] = 0 ( x 2 + 1 + 2 x + 2 ) ( x 2 + 1 − 2 x − 2 ) = 0 ( x 2 + 2 x + ( 1 + 2 ) ) ( x 2 − 2 x + ( 1 − 2 ) ) = 0 <

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

6th Edition

ISBN: 9781524908102

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 1.2, Problem 51PS

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To find:all roots of the quartic equation x4−4x=1 .

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Chapter 1.2, Problem 50PS

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Chapter 1 Solutions

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

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all roots of the quartic equation x 4 − 4 x = 1 .

Solution Summary: The author calculates the quartic equation x4-4x=1, which is a polynomial of degree 4 and has 4 roots.

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To find:all roots of the quartic equation x4−4x=1 .

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Chapter 1 Solutions