Metode Bairstrow

mencari akar persamaan
c. y = f(x) = x^4 + 4x³ + 21x² + 4x + 20 = 0
(x² - rx + s) (b1x² + b2x + b3)

Mencari b1, b2, b3, r, s dalam table :



r = 0 ; s = -1 ; b1 = 1 ; b2 = 4 ; b3 = 20

(x² + 1) (x² + 4x + 20)

x1,2 = -b±√b²-4ac
2a

= - 4±√(4)²-4*1*20
2*1

= - 4 ± 8 i
2

X1 = -2 + 4i
X2 = -2 - 4i

Metode faktorisasi p5

mencari akar persamaan
b. y = f(x) = X5-3.5x4-8.5x3+29.75x2+14.0625x-49.21875=0
(x+a)(x^2+b1x+b0)(x^2+4x+c0)

Mencari a, b1, b0, c1, c0 dalam table :


iterasi b0 b1 a c1 c0
1 0 0 0 -3.5 -8.5
2 -1.6544118 -2.81877 -3.5000037 2.81877 1.0998743
3 12.785555 -32.766983 3.4999988 25.767001 847.52172
4 0.1249966 -0.0347729 -0.4645987 -3.00063 -10.139576
5 -2.5390737 -1.7285583 -1.9117578 0.14032 -8.7547122
6 -3.2866953 -0.9316686 -1.7105137 -0.85782 -9.0734457
7 -3.4303003 -1.4754887 -1.5813348 -0.44318 -8.5165734
8 -1.6544118 -1.2654929 -3.4931763 1.25867 -5.2765792
9 -2.6483628 -0.6269518 -3.5220768 0.64903 -5.3669631
10 -2.6767827 -0.3397704 -3.4259998 0.26577 -5.9865113


=(x-3.4259998)(x^2-0.3397704-2.6767827)(x^2+0.26577x-5.9865113)
x1 =3.4259998

(x^2-0.3397704-2.6767827)
x1,2 = -b±√b²-4ac
2a
= --0.3397704±√(-0.3397704)²-4*1*-2.6767827
2*1
=0.3397704±3.289768
Jadi 2
x2= 1.8147692
x3= -1.4749988

(x^2+0.26577x-5.9865113)
x4,5 = -b±√b²-4ac
2a
= -0.26577±√(0.26577)²-4*1*-5.9865113
2*1
= -0.26577±4.90068
2
Jadi
x4 = 1.1587275
x5 = -5.166257

Metode Faktorisasi p4

mencari akar persamaan
a. y = f(x) = x^4 - x³ - 7x² + x + 6 = 0
(x² + b1x + b0) (x² + a1x + a0)

Mencari b1, b0, a1, a0 dalam table :




bo = -1 b1 = 0 a1 = -1 a0 = -6

(x² + b1x + b0) (x² + a1x + a0)
(x² - 1) (x² - x – 6)
(x + 1) (x – 1) (x – 3) (x + 2)

Jadi x^4 - x³ - 7x² + x + 6 = 0
Menghasilkan akar-akar
x 1 = -1 x2 = 1 x3 = 3 x4 = -2

Metode Newton Raphson

mencari akar persamaan
b. y = f(x) = x³ + x² – 3x + 3 = 0
f ‘ (x) = 3x² + 2x - 3 = 0
f “ (x) = 6x + 2 = 0

misal x1 = -1, mencari x2 = x1 – (f(x1)/f’(x1))

mencari x2, x3, x4 dst dalam table :





Jadi y = f(x) = x³ + x² – 3x + 3 = 0
menghasilkan
x = 0,491669
dengan error sebesar -1,29144

Metode Biseksi

mencari akar persamaan
b. y = f(x) = x³ – x² – 2x + 1 = 0

menentukan x1 = 0 dan x2 = 1
f(x1) = f(0) = 0³ – 0² + 2(0) + 1 = 1
f(x2) = f(1) = 1³ – 1² + 2(1) + 1 = -1

f(x1).f(x2) = 1 (-1) < 0
- 1 < 0

mencari x3,x4,x5 dst dalam tabel


METODE BISEKSI




jadi y = x³ – x² – 2x + 1 = 0
menghasilkan
x = 0.500977
dengan error
sebesar -0.1272

Metode Tabulasi

mencari akar persamaan

b. y = f(x) = 2 – 3x + sin x = 0

tentukan x1=0 dan x2=1

f(x1)=f(0)=2-3(0)+sin(0)=2

f(x2)=f(1)=2-3(1)+sin(1)=-0.98254

f(x1).f(x2)=2(-0.98254)<0

membuat tabel fungsi disekitar 0 dan 1

METODE TABULASI
x	f(x)	AE
0.0	2	1.4912736
0.1	1.7017453	1.1930189
0.2	1.4034907	0.8947642
0.3	1.105236	0.5965096
0.4	0.8069813	0.2982549
0.5	0.5087265	1.329E-07
0.6	0.2104718	0.2982546
0.7	-0.087783	0.5965094
0.8	-0.3860378	0.8947642
0.9	-0.6842927	1.1930191
1.0	-0.9825476	1.491274
x	0.5087264
x	f(x)	AE
0.60	0.2104718	0.1491274
0.61	0.1806463	0.1193019
0.62	0.1508208	0.0894764
0.63	0.1209954	0.059651
0.64	0.0911699	0.0298255
0.65	0.0613444	1.728E-09
0.66	0.0315189	0.0298255
0.67	0.0016934	0.059651
0.68	-0.028132	0.0894764
0.69	-0.0579575	0.1193019
0.70	-0.087783	0.1491274
x	0.0613444
x	f(x)	AE
0.670	0.0016934	0.0115259
0.671	-0.0012891	0.0145084
0.672	-0.0042717	0.017491
0.673	-0.0072542	0.0204735
0.674	-0.0102368	0.0234561
0.675	-0.0132193	0.0264386
0.676	-0.0162018	0.0294211
0.677	-0.0191844	0.0324037
0.678	-0.0221669	0.0353862
0.679	-0.0251495	0.0383688
0.680	-0.028132	0.0413513
x	-0.0132193
x	f(x)	AE
0.6700	0.0016934	0.0016404
0.6701	0.0013952	0.0013421
0.6702	0.0010969	0.0010439
0.6703	0.0007987	0.0007456
0.6704	0.0005004	0.0004474
0.6705	0.0002022	0.0001491
0.6706	-9.609E-05	0.0001491
0.6707	-0.0003943	0.0004474
0.6708	-0.0006926	0.0007456
0.6709	-0.0009909	0.0010439
0.6710	-0.0012891	0.0013421
x	5.304E-05
x	f(x)	AE
0.67050	0.0002022	0.000164
0.67051	0.0001723	0.0001342
0.67052	0.0001425	0.0001044
0.67053	0.0001127	7.456E-05
0.67054	8.286E-05	4.474E-05
0.67055	5.304E-05	1.491E-05
0.67056	2.321E-05	1.491E-05
0.67057	-6.613E-06	4.474E-05
0.67058	-3.644E-05	7.456E-05
0.67059	-6.626E-05	0.0001044
0.67060	-9.609E-05	0.0001342
x	3.813E-05
x	f(x)	AE
0.670560	2.321E-05	1.64E-05
0.670561	2.023E-05	1.342E-05
0.670562	1.725E-05	1.044E-05
0.670563	1.427E-05	7.456E-06
0.670564	1.128E-05	4.474E-06
0.670565	8.3E-06	1.491E-06
0.670566	5.317E-06	1.491E-06
0.670567	2.335E-06	4.474E-06
0.670568	-6.477E-07	7.456E-06
0.670569	-3.63E-06	1.044E-05
0.670570	-6.613E-06	1.342E-05
x	6.809E-06
x	f(x)	AE
0.6705670	2.335E-06	1.64E-06
0.6705671	2.037E-06	1.342E-06
0.6705672	1.738E-06	1.044E-06
0.6705673	1.44E-06	7.456E-07
0.6705674	1.142E-06	4.474E-07
0.6705675	8.436E-07	1.491E-07
0.6705676	5.453E-07	1.491E-07
0.6705677	2.47E-07	4.474E-07
0.6705678	-5.121E-08	7.456E-07
0.6705679	-3.495E-07	1.044E-06
0.6705680	-6.477E-07	1.342E-06
x	6.944E-07

jadi y = f(x) = 2 – 3x + sinx = 0

akan menghasilkan

x = 0.6705677

dengan error sebesar 4.474 x 10-7