To determine To solve: The equation x 2 3 + 9 x 1 3 + 14 = 0 analytically. Explanation Formula used: Quadratic formula: The solution of the quadratic equation a x 2 + b x + c = 0 is calculated using the quadratic formula as shown below: x = − b ± b 2 − 4 a c 2 a Calculation: The given equation can be solved by converting it into quadratic form. Assume u = x 1 3 . Then the equation becomes u 2 + 9 u + 14 = 0 . Solve the above equation further by using the quadratic formula as follows. u = − ( 9 ) ± ( 9 ) 2 − 4 × 1 × ( 14 ) 2 × 1 = − 9 ± 81 − 56 2 = − 9 ± 25 2 = − 9 ± 5 2 Now there are two solutions for the value of u . The first solution is calculated as: u = − 9 − 5 2 = − 14 2 = − 7 The second solution is calculated as: u = − 9 + 5 2 = − 4 2 = − 2 Hence, the values of u are u = − 7 and u = − 2 . Substitute u = − 7 in x 1 3 = u . x 1 3 = − 7 x = ( − 7 ) 3 x = − 343 Again, substitute u = − 2 in x 1 3 = u . x 1 3 = − 2 x = ( − 2 ) 3 x = − 8 Hence, the values of x are x = − 343 and x = − 8

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ISBN: 9780134696522

Author: HORNSBY, John, Lial, Margaret L., Rockswold, Gary K.

Publisher: Pearson Education, Inc.,

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Chapter 5.5, Problem 30E

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To solve: The equation x23+9x13+14=0 analytically.

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Chapter 5.5, Problem 29E

Chapter 5.5, Problem 31E

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Graphical Approach To College Algebra

Ch. 5.1 - Provide a short answer to each question. Do not...Ch. 5.1 - Prob. 2E Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - Prob. 5E Ch. 5.1 - Provide a short answer to each question. Do not...Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - Prob. 9E Ch. 5.1 - Prob. 10E

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