Solve Q14, 15, 16 explaining detailly each step 9. Given that a root of 2x3 – x² + mx – 4 = (0 lies in the interval (1, 2) the range of values of mis:A. –8 < m< 3 B. -4< m < 3 C. m<-4 or m< 3 D. m<-4 or m2 310. If a real negative root of: x + 3x² – x – 8 = 0 lies in the interval (a, a + 1), the value of theinteger a isA. -2B. -1 C. 0 D. 13-x7and starting with311. By performing one interaction using the procedure defined by Xn+1X = 0.8, an approximation to the root of x'+ 3x - 3 = 0 isA. 0.163.3B. 0.488C. 0.512 D. 0.17112. A first approximation to the real root of the equation: x³ + x² – 5x – 1 = 0 is 2. A secondapproximation to the root of the equation, using the Newton-Raphson's method, is19A.1121В.1118C.1123D.1113. Starting with x = 1.5 and performing one interaction using the procedure defined by x1=%3D3x0-13an approximate value of the root of x'- 3x +1 = 0 is7A.27С.2B. V3D.314. Starting with the initial value x =and using Newton-Raphson procedure once, an2approximation to the root of x – 2 = 0 is17В. -—12|1А.1219С.-1211D.-8.15. It is given that x = a is a first approximation of the equation f(x) = 0, wheref (x) = x3 – 3x + 1. Using the Newton-Raphson method, a second approximation of theroot is2a3-1D.За3—1С.3 (а2-1)216. By writing x = 2 + h, and neglecting terms in h and h' an approximate vaiue of the root of2a3-1A.3(а2-1)2a3+1B.3 (а?-1)3 (а3-1)x3+ x - 11 = 0 in the interval 2

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Answered: 14. Starting with the initial value x =…

14. Starting with the initial value x = and using Newton-Raphson procedure once, an approximation to the root of x- 2=0 is 17 В. 12 1 A. 12 19 C. -- 12 11 D. 8.

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter8: Polynomials

Section8.4: Zeros Of A Polynomial

Problem 31E

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Equations and Inequations

Equations and inequalities describe the relationship between two mathematical expressions.

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A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.

Question

Solve Q14, 15, 16 explaining detailly each step

9. Given that a root of 2x3 – x² + mx – 4 = (0 lies in the interval (1, 2) the range of values of m
is:
A. –8 < m< 3 B. -4< m < 3 C. m<-4 or m< 3 D. m<-4 or m2 3
10. If a real negative root of: x + 3x² – x – 8 = 0 lies in the interval (a, a + 1), the value of the
integer a is
A. -2
B. -1 C. 0 D. 1
3-x7
and starting with
3
11. By performing one interaction using the procedure defined by Xn+1
X = 0.8, an approximation to the root of x'+ 3x - 3 = 0 is
A. 0.163
.3
B. 0.488
C. 0.512 D. 0.171
12. A first approximation to the real root of the equation: x³ + x² – 5x – 1 = 0 is 2. A second
approximation to the root of the equation, using the Newton-Raphson's method, is
19
A.
11
21
В.
11
18
C.
11
23
D.
11
13. Starting with x = 1.5 and performing one interaction using the procedure defined by x1=
%3D
3x0-1
3
an approximate value of the root of x'- 3x +1 = 0 is
7
A.
2
7
С.
2
B. V3
D.
3
14. Starting with the initial value x =
and using Newton-Raphson procedure once, an
2
approximation to the root of x – 2 = 0 is
17
В. -—
12
|
1
А.
12
19
С.-
12
11
D.-
8.
15. It is given that x = a is a first approximation of the equation f(x) = 0, where
f (x) = x3 – 3x + 1. Using the Newton-Raphson method, a second approximation of the
root is
2a3-1
D.
За3—1
С.
3 (а2-1)
2
16. By writing x = 2 + h, and neglecting terms in h and h' an approximate vaiue of the root of
2a3-1
A.
3(а2-1)
2a3+1
B.
3 (а?-1)
3 (а3-1)
x3+ x - 11 = 0 in the interval 2<x< 3 is
12
13
5
В.-
2
27
C.=
13
А.
13

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