Explanation Formula used: Quadratic formula : For a quadratic equation of the form a x 3 + b x + c = 0 the solutions are x = − b ± b 2 − 4 a c 2 a . Procedure used: Solving radical equations: 1. Isolate a single radical on one side of the equation. 2. Square both sides of the equation. (Or raise both sides to a power that is equal to the index of the radical.). 3. If a radica1 remains, repeat steps 1 and 2. 4. Solve the resulting equation. 5. All solutions must be checked in the original equation, and only those that satisfy the original equation are actual solutions. Calculation: By using above procedure, the given equation can be solved as follows. 2 x 2 3 + 5 x 1 3 − 12 = 0 2 x 2 3 + 5 x 1 3 = 12 [ ∵ Isolating the radical ] Let u = x 3 . Then the equation 2 x 2 3 + 5 x 1 3 = 12 becomes 2 u 2 + 5 u − 12 = 0 . Substitute a = 2 , b = 5 and c = − 12 in the quadratic formula to find the value of u . u = − 5 ± ( 5 ) 2 − 4 ( 2 ) ( − 12 ) 2 ( 2 ) = − 5 ± 25 + 96 4 = − 5 ± 121 4 = − 5 ± 11 4 On further simplification, u = − 5 + 11 4 or − 5 − 11 4 = 6 4 or − 16 4 = 3 2 or − 4 Here, the values of u are u = 3 2 and u = − 4 . Substitute u = 3 2 in u = x 3 to find the value of x . 3 2 = x 3 x = ( 3 2 ) 3 x = 27 8 Here, the value of x is x = 27 8

College Algebra in Context with Applications for the Managerial, Life, and Social Sciences (5th Edition)

5th Edition

ISBN: 9780134179025

Author: Ronald J. Harshbarger, Lisa S. Yocco

Publisher: PEARSON

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Chapter 4.4, Problem 19E

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To solve: The equation 2x23+5x13−12=0 algebraically and check the solution graphically or by substitution.

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Chapter 4.4, Problem 18E

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College Algebra in Context with Applications for the Managerial, Life, and Social Sciences (5th Edition)

Ch. 4.1 - In Exercises 116, (a) sketch the graph of each...Ch. 4.1 - In Exercises 116, (a) sketch the graph of each...Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - In Exercises 116, (a) sketch the graph of each...Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

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The equation 2 x 2 3 + 5 x 1 3 − 12 = 0 algebraically and check the solution graphically or by substitution.

Solution Summary: The author explains how to solve the equation 2x23+5 if the solution is graphically or by substitution.

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To solve: The equation 2x23+5x13−12=0 algebraically and check the solution graphically or by substitution.

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