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

7th Edition

ISBN: 9781337291248

Author: Waner, Stefan.

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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For Exercises 115–120, factor the expressions over the set of complex numbers. For assistance, consider these examples. • In Section R.3 we saw that some expressions factor over the set of integers. For example: x - 4 = (x + 2)(x – 2). • Some expressions factor over the set of irrational numbers. For example: - 5 = (x + V5)(x – V5). To factor an expression such as x + 4, we need to factor over the set of complex numbers. For example, verify that x + 4 = (x + 2i)(x – 2i). 115. а. х - 9 116. а. х? - 100 117. а. х - 64 b. x + 9 b. + 100 b. x + 64 118. а. х — 25 119. а. х— 3 120. а. х — 11 b. x + 25 b. x + 3 b. x + 11

For Exercises 19–26, simplify each expression and write the result in standard form, a + bi. 8 + 3i 19. 4 + 5i 20. -4 - 6i 21. 9 - 15i 22. 14 6. -2 -3 -18 + V-48 23. - 20 + V-50 14 - V-98 25. - 10 + V-125 24. 26. 4 10 -7

Use the Gauss-Jordan technique to solve3x1 – 0.1x2 – 0.2x3 = 7.850.1x1 + 7x2 – 0.3x3 = -19.30.3x1 – 0.2x2 + 10x3 = 71.4 Please solve in excel data tabling, thanks in advance.

Chapter 0 Solutions

Applied Calculus

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Ch. 0.1 - Prob. 11E Ch. 0.1 - Prob. 12E Ch. 0.1 - Prob. 13E Ch. 0.1 - Calculate each expression in Exercises 124, giving...Ch. 0.1 - Prob. 15E Ch. 0.1 - Prob. 16E Ch. 0.1 - Calculate each expression in Exercises 124, giving...Ch. 0.1 - Prob. 18E Ch. 0.1 - Prob. 19E Ch. 0.1 - Prob. 20E Ch. 0.1 - Prob. 21E Ch. 0.1 - Prob. 22E Ch. 0.1 - Prob. 23E Ch. 0.1 - Prob. 24E Ch. 0.1 - Prob. 25E Ch. 0.1 - Prob. 26E Ch. 0.1 - Prob. 27E Ch. 0.1 - Prob. 28E Ch. 0.1 - Prob. 29E Ch. 0.1 - Prob. 30E Ch. 0.1 - Prob. 31E Ch. 0.1 - Prob. 32E Ch. 0.1 - Prob. 33E Ch. 0.1 - Prob. 34E Ch. 0.1 - Prob. 35E Ch. 0.1 - Prob. 36E Ch. 0.1 - Convert each expression in Exercises 25-50 into...Ch. 0.1 - Prob. 38E Ch. 0.1 - Prob. 39E Ch. 0.1 - Prob. 40E Ch. 0.1 - Prob. 41E Ch. 0.1 - Prob. 42E Ch. 0.1 - Prob. 43E Ch. 0.1 - Prob. 44E Ch. 0.1 - Prob. 45E Ch. 0.1 - Prob. 46E Ch. 0.1 - Prob. 47E Ch. 0.1 - Prob. 48E Ch. 0.1 - Prob. 49E Ch. 0.1 - Prob. 50E Ch. 0.2 - Evaluate the expression sin Exercises 116. 33 Ch. 0.2 - Prob. 2E Ch. 0.2 - Evaluate the expression sin Exercises 116. 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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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Chapter 0 Solutions