To determine To solve: The equation x 2 3 − 2 x 1 3 − 15 = 0 algebraically. Explanation Definition used: General quadratic equation : “Given that a , b , and c are constant ( a ≠ 0 ) , the equation a x 2 + b x 2 + c = 0 is called the general quadratic equation in x ”. Calculation: Consider the equation x 2 3 − 2 x 1 3 − 15 = 0 . In order to find the value of x , first rewrite the equation in quadratic form. Substitute y = x 1 3 in x 2 3 − 2 x 1 3 − 15 = 0 and simplify the equation as follows. y 2 − 2 y − 15 = 0 y 2 + 3 y − 5 y + 15 = 0 y ( y + 3 ) − 5 ( y + 3 ) = 0 ( y − 5 ) ( y + 3 ) = 0 On further simplification, y − 5 = 0 or y + 3 = 0 y</

Mylab Math With Pearson Etext -- 18 Week Standalone Access Card -- For Basic Technical Mathematics With Calculus

11th Edition

ISBN: 9780135902912

Author: Allyn J. Washington

Publisher: PEARSON

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Chapter 14.3, Problem 13E

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To solve: The equation x23−2x13−15=0 algebraically.

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Chapter 14.3, Problem 12E

Chapter 14.3, Problem 14E

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Chapter 14 Solutions

Mylab Math With Pearson Etext -- 18 Week Standalone Access Card -- For Basic Technical Mathematics With Calculus

Ch. 14.1 - In Example 7, determine how many times the rocket...Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E Ch. 14.1 - Prob. 6E Ch. 14.1 - Prob. 7E Ch. 14.1 - Prob. 8E Ch. 14.1 - Prob. 9E

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