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