Solution 3.8.8.1
The factorization of the denominator is
s = ;
1
2
1
2
q
2
1
;4
0
:
Now
2
1
=
1
1
+
1
t
2
+
1
R
2
C
2
=
1
2
1
+
1
2
2
+
2
1
2
+
2
1
R
2
C
+
2
2
R
2
C
+
1
(R
2
C)
2
Then
2
1
;4
0
=
1
2
1
+
1
2
2
+
2
1
2
+
2
1
R
2
C
+
2
2
R
2
C
+
1
(R
2
C)
2
;
4
1
2
=
1
2
1
+
1
2
2
;
2
1
2
+
2
1
R
2
C
+
2
2
R
2
C
+
1
(R
2
C)
2
=
1
1
;
1
2
2
+
2
1
R
2
C
+
2
2
R
2
C
+
1
(R
2
C)
2
But all the terms in this nal expression are positiveandhence
2
1
;4
0
> 0;;
and the tworoots will always be real.
1