Explanation Procedure used: 1 . Find an equivalent inequality with P ( x ) on one side and 0 on the other. 2. Change the inequality symbol to an equals sign and solve the related equation; that is, solve P ( x ) = 0 . 3. Use the solutions to divide the x axis into intervals. Then select a test value from each interval and determine the sign of the polynomial on the interval. 4. Determine the intervals for which the inequality is satisfied and write intervals notation or set builder notation for the solution set. Include the endpoints of the intervals in the solution set if the inequality symbol is ≤ or ≥ . Calculation: The number of diagonals D for a polygon of n sides is given by, D ( n ) = n ( n − 3 ) 2 . The number of diagonals lies between 27 and 230. That is, 27 ≤ D ≤ 230 . Substitute the value of D in the above inequality and simplify. 27 ≤ n ( n − 3 ) 2 ≤ 230 54 ≤ ( n 2 − 3 n ) ≤ 460 Use the steps from 1 to 5 to find the solution set of the inequality. Simplify the above polynomial inequality 54 ≤ ( n 2 − 3 n ) ≤ 460 as follows. Case 1: First consider the inequality 54 ≤ n 2 − 3 n . An equivalent inequality with 0 on one side is n 2 − 3 n − 54 ≥ 0 . Change the inequality sign by equal sign to form related equation as n 2 − 3 n − 54 = 0 . Find the value of n from the equation n 2 − 3 n − 54 = 0 . n 2 − 3 n − 54 = 0 n 2 − 9 n + 6 n − 54 = 0 n ( n − 9 ) + 6 ( n − 9 ) = 0 ( n − 9 ) ( n + 6 ) = 0 On further simplification, n − 9 = 0 or n + 6 = 0 n = 9 or n = − 6 Since the number of sides cannot be negative, the possible value of n is 9. The value of n forms the interval ( − ∞ , 9 ) , and ( 9 , ∞ ) . Find the sign of the polynomial function on each interval by selecting test value as follows. Let the polynomial function is denoted by g ( n ) = n 2 − 3 n − 54 . Obtain the sign of polynomial g ( n ) = n 2 − 3 n − 54 at n = 0 on the interval ( − ∞ , 9 ) . g ( 0 ) = ( 0 ) 2 − 3 ( 0 ) − 54 = 0 − 0 − 54 = − 54 < 0 Obtain the sign of polynomial g ( n ) = n 2 − 3 n − 54 at n = 10 on the interval ( 9 , ∞ ) . g ( 10 ) = ( 10 ) 2 − 3 ( 10 ) − 54 = 100 − 30 − 54 = 16 > 0 From the above calculation, it is clear that the inequality 54 ≤ n 2 − 3 n satisfy on the interval ( 9 , ∞ )

College Algebra: Graphs and Models (6th Edition)

6th Edition

ISBN: 9780134179032

Author: Marvin L. Bittinger, Judith A. Beecher, David J. Ellenbogen, Judith A. Penna

Publisher: PEARSON

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Chapter 4.6, Problem 81E

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The number of sides n in a polygon.

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Chapter 4.6, Problem 80E

Chapter 4.6, Problem 82E

Chapter 4 Solutions

College Algebra: Graphs and Models (6th Edition)

Ch. 4.1 - Determine the leading term, the leading...Ch. 4.1 - Prob. 2E Ch. 4.1 - Determine the leading term, the leading...Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Determine the leading term, the leading...Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Determine the leading term, the leading...Ch. 4.1 - Prob. 10E

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The number of sides n in a polygon .

Solution Summary: The author calculates the number of diagonals D for a polygon of n sides by dividing the x axis into intervals.

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The number of sides n in a polygon.

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