To determine The solutions for the given equation. Explanation Given: The given equation is 5 x 2 − 3 x − 3 = 0 , [ 1 , 2 ] Calculation: Newton’s approximation method can be expressed as, C n + 1 = C n − f ( C n ) f ′ ( C n ) . Here, C n is current approximation value and C n + 1 upcoming approximation value. Consider f ( x ) = 5 x 2 − 3 x − 3 (1) Now, differentiate the above expression with respect to x . f ′ ( x ) = 10 x − 3 (2) Substitute x = 1 in equation (1), f ( 1 ) = 5 ( 1 ) 2 − 3 ⋅ 1 − 3 = − 1 Substitute x = 2 in equation (1), f ( 2 ) = 5 ( 2 ) 2 − 3 ⋅ 2 − 3 = 20 − 6 − 3 = 11 Substitute x = 1 in equation (2), f ′ ( x ) = 10 ⋅ 1 − 3 = 7 Check the values ( f ( 1 ) < 0 ) and ( f ( 2 ) > 0 ) , the values of f ( 1 ) and f ( 2 ) are in opposite sign so there is a solution for the equation in interval ( 1 , 2 ) . Consider an initial value C 1 = 1 Substitute the value of f ( C 1 ) , f ′ ( C 1 ) , C 1 and n = 1 for first iteration in Newton’s formula. C 1 + 1 = C 1 − f ( C 1 ) f ′ ( C 1 ) C 2 = C 1 − f ( 1 ) f ′ ( 1 ) = 1 − ( − 1 ) 7 = 1

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Chapter 12.6, Problem 1E

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The solutions for the given equation.

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Chapter 12.6, Problem 2WE

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