Explanation Given Information: The provided equation is, 3 x 2 = 1 Formula used: Steps to solve the equation of type x 2 = a : Step 1: Write the provided equation in standard form, that is x 2 − a = 0 . Step 2: Make factors. Step 3: Use zero product rule on factors as: If ( x + p ) ( x + q ) = 0 Then, ( x + p ) = 0 or ( x + q ) = 0 step 4: Simplify to isolate the variable. Step 5: Check the solution by substituting it in the original equation. Calculation: Consider the provided equation, 3 x 2 = 1 Divide both sides by 3: 3 x 2 3 = 1 3 x 2 = 1 3 Subtract from both sides: x 2 − 1 3 = 1 3 − 1 3 x 2 − 1 3 = 0 x 2 − ( 1 3 ) 2 = 0 As, x 2 − a 2 = ( x + a ) ( x − a ) x 2 − ( 1 3 ) 2 = 0 is of type x 2 − a 2 Thus, factors of the equation x 2 − ( 1 3 ) 2 = 0 are: x 2 − ( 1 3 ) 2 = 0 ( x − 1 3 ) ( x + 1 3 ) = 0 As, zero product rule is: If ( x + p ) ( x + q ) = 0 Then, ( x + p ) = 0 or ( x + q ) = 0 So, apply zero product rule on ( x − 1 3 ) ( x + 1 3 ) = 0 : Thus, ( x − 1 3 ) = 0 or ( x + 1 3 ) = 0 If, x + 1 3 = 0 Then, x

Elementary and Intermediate Algebra: Concepts and Applications (7th Edition)

7th Edition

ISBN: 9780134462707

Author: Marvin L. Bittinger, David J. Ellenbogen, Barbara L. Johnson

Publisher: PEARSON

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Chapter 11.1, Problem 3YT

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Chapter 11 Solutions

Elementary and Intermediate Algebra: Concepts and Applications (7th Edition)

Ch. 11.1 - Solve: 64=y2.Ch. 11.1 - Solve: t2=10. Give exact solutions and...Ch. 11.1 - Prob. 3YT Ch. 11.1 - Prob. 4YT Ch. 11.1 - Prob. 5YT Ch. 11.1 - 6. Solve: . Ch. 11.1 - Prob. 7YT Ch. 11.1 - Prob. 8YT Ch. 11.1 - Prob. 9YT Ch. 11.1 - Prob. 10YT

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The solution of the equation 3 x 2 = 1 .

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Chapter 11 Solutions