Explanation Formula used: Newton’s approximation of x n is given by, x n = x n − 1 − f ( x n ) f ′ ( x n ) . Calculation: The given quadratic equation is, f ( x ) = 2 x 2 − x − 2 … … ( 1 ) Find the values of the quadratic equation at the boundary points. Substitute x = 1 in equation (1). f ( 1 ) = 2 ( 1 ) 2 − ( 1 ) − 2 = 2 − 1 − 2 = − 1 Substitute x = 2 in equation (1). f ( 2 ) = 2 ( 2 ) 2 − ( 2 ) − 2 = 8 − 2 − 2 = 4 Note the value of x 1 lies between x = 1 and x = 2 , hence use the value x 1 = 1.5 . Step 1: Differentiate equation (1) with respect to x. f ′ ( x ) = 4 x − 1 … … ( 2 ) Substitute x 1 = 1.5 in equation (1). f ( 1.5 ) = 2 ( 1.5 ) 2 − ( 1.5 ) − 2 = 4.5 − 1.5 − 2 = 1 Substitute x 1 = 1.5 in equation (2). f ′ ( 1.5 ) = 4 ( 1.5 ) − 1 = 6 − 1 = 5 By Newton’s approximation, x 2 = x 1 − f ( x 1 ) f ′ ( x 1 ) x 2 = 1

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ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Chapter 24.2, Problem 2E

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The root of the quadratic equation 2x2−x−2 between x=1 and x=2 by Newton’s method and compare this root with the root obtained by the quadratic formula.

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Chapter 24.2, Problem 1E

Chapter 24.2, Problem 3E

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Chapter 24 Solutions

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The root of the quadratic equation 2 x 2 − x − 2 between x = 1 and x = 2 by Newton’s method and compare this root with the root obtained by the quadratic formula.

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The root of the quadratic equation 2x2−x−2 between x=1 and x=2 by Newton’s method and compare this root with the root obtained by the quadratic formula.

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Chapter 24 Solutions