Explanation Given Information: The provided polynomial equation is x 4 + x 2 − 1 = 0 . Formula used: The value of x in the equation a x 2 + b x + c = 0 can be given by the quadratic formula , x = − b ± b 2 − 4 a c 2 a where a , b and c are real numbers with a ≠ 0 . Calculation: Consider the polynomial equation x 4 + x 2 − 1 = 0 . The above polynomial equation can be written as ( x 2 ) 2 + x 2 − 1 = 0 Let x 2 = t put in the polynomial equation ( x 2 ) 2 + x 2 − 1 = 0 t 2 + t − 1 = 0 Compare the above equation with the equation a x 2 + b x + c = 0 , a = 1 , b = 1 and c = − 1 Put the values of a , b , c in the quadratic formula x = − b ± b 2 − 4 a c 2 a . t = − 1 ± ( 1 ) 2 − 4 ⋅ 1 ( − 1 ) 2 ⋅ 1 = − 1 ± 1 + 4 2 = − 1 ± 5 2 Further simplify, Put − 1 ± 5 2 in x 2 = t , x 2 = t = − 1 ± 5 2 Take square root from both sides, x = ± − 1 ± 5 2 Check: For x = − 1 ± 5 2 Substitute x = − 1 ± 5 2 in the equation x 4 + x 2 − 1 = 0 then simplify, ( − 1 ± 5 2 ) 4 + ( − 1 ± 5 2 ) 2 − 1 = ? 0 ( − 1 ± 5 2 ) 2 + ( − 1 ± 5 2 ) − 1 = ? 0 Taking positive ( + ) sign from the above equation then simplify, ( − 1 + 5 2 ) 2 + ( − 1 + 5 2 ) − 1 = ? 0 1 + 5 − 2 5 4 + − 1 + 5 2 − 1 = ? 0 6 − 2 5 − 2 + 2 5 − 4 4 = ? 0 0 = 0 Which is true

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ISBN: 9781337280426

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Chapter 0.5, Problem 29E

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To calculate: The all real solutions in the polynomial equation x4+x2−1=0.

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Chapter 0.5, Problem 28E

Chapter 0.5, Problem 30E

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The all real solutions in the polynomial equation x 4 + x 2 − 1 = 0 .

Solution Summary: The author calculates the polynomial equation x4+2-1=0 using a quadratic formula.

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To calculate: The all real solutions in the polynomial equation x4+x2−1=0.

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