Civil Engineering Reference
In-Depth Information
Table 1.29.
Example 1.17 Bairstow's method
0.9898
1
24
−5
−2
0.990
0.012
−3.969
5.9816
5.982
−23.987
0.9898
1
0.012
0.025
−4.010
0.990
−2.990
5.9816
5.982
1
−
3.020
3.004
0 025
.
−
3 020
.
(
)()
−−
(
)
(
)
0 012
.
.
1
0 025 1
.
3 020
.
0 012
.
∆
u
=
=
(
)()
−−
(
)
−
(
)
3 004
−
3 020
.
3 004 1
.
3 020
.
3 020
.
−
3 020
.
1
0 061
6 116
.
=
−
=−
0 0100
.
.
3 004 0 025
3 020 0 012
3 004
.
.
=
(
)(
)
−
(
)
−
(
)
−
.
.
3 004 0 012
.
.
0 025
.
3 020
.
∆
v
=
(
)()
−−
(
)
−
(
)
.
−
3 020
.
3 004 1
.
3 020
.
3 020
.
−
3 020
.
1
0 111
611
.
.
=
−
=− .
0 0181
6
uu u
2
=+ =−
∆
0 9898 0 0100
.
−
.
= −
0 9998
.
vv v
2
=+ =−
∆
5 9816 0 0181
.
−
.
= −
5 9997
.
It appears the values are
u = −
1 and
v = −
6. Repeat the process using the
revised values for
u
and
v
as shown in Table 1.30.
Table 1.30.
Example 1.17 Bairstow's method
1
1
−
5
−
2
24
1
−
4
0
6
6
−
24
1
−
4
0
0
Since the remainders of the first division
b
n-
1
and
b
n
are both zero,
u = −
1
and
v = −
6 are the coefficients of the root quadratic. Substitute them into
the quadratic equation to find the roots.
