Explanation Given information: The inequality is x 4 − 7 x 2 − 18 < 0 . Calculation: The given inequality is x 4 − 7 x 2 − 18 < 0 . Consider P ( x ) = x 4 − 7 x 2 − 18 , then factorize the polynomial P ( x ) as, P ( x ) = x 4 − 7 x 2 − 18 = x 4 − 9 x 2 + 2 x 2 − 18 = x 2 ( x 2 − 9 ) + 2 ( x 2 − 9 ) Further calculate the above expression as, P ( x ) = ( x 2 − 9 ) ( x 2 + 2 ) = ( x + 3 ) ( x − 3 ) ( x 2 + 2 ) Therefore, the factorized form of the polynomial is P ( x ) = ( x + 3 ) ( x − 3 ) ( x 2 + 2 ) . The zeros of the polynomial P ( x ) is calculated as, P ( x ) = 0 ( x + 3 ) ( x − 3 ) ( x 2 + 2 ) = 0 x + 3 = 0 or x − 3 = 0 or x 2 + 2 = 0 x = − 3 or x = 3 or x 2 + 2 = 0 Therefore, the real zeros of the polynomial P ( x ) are − 3 and 3 as the factor x 2 + 2 has no real zeros. Hence, the intervals determined by the zeros of the polynomial are ( − ∞ , − 3 ) , ( − 3 , 3 ) and ( 3 , ∞ ) . The factor x 2 + 2 is positive everywhere. The sign of eachremaining factor, in each interval are shown in Table 1. Sign of ( − ∞ , − 3 ) ( − 3 , 3 ) ( 3 , ∞ ) ( x + 3 ) − + + ( x − 3 ) − − + P ( x ) = ( x + 3 ) ( x − 3 ) ( x 2 + 2 ) + − + Table 1 From the Table 1, it is shown that the inequality P ( x ) < 0 satisfies in the interval ( − 3 , 3 ) . The end points of the intervals ( − ∞ , − 3 ) , ( − 3 , 3 ) and ( 3 , ∞ ) , are − 3 and 3 . To check whether the end points satisfies the inequality P ( x ) < 0 or not,substitute the end points for x in the inequality P ( x ) < 0 respectively

7th Edition

ISBN: 9781305115545

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: Cengage Learning

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Chapter 3.7, Problem 11E

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The inequality x4−7x2−18<0.

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Chapter 3.7, Problem 10E

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