357 



so that the equation has six imaginary roots. A.s the last term of the 

 equation is negative, the two remaining roots are real — one positive, 

 and the other negative. 



It may be observed here that when the extreme terms of any triad 

 have nnlike signs, as is the case with the third, fourth, and sixth of 

 the triads above, it may always be passed over ; the corresponding sign 

 in the row of signs being written +, to imply that the triad in ques- 

 tion fails to satisfy the condition of imaginary roots. 



It may be remarked, too, that the accurate calculation of the two 

 members of the inequality appealed to is but seldom necessary. Which 

 of these two members is in excess of the other may, in most instances, 

 be ascertained at a glance. 



In the foregoing discussion, the case in which the sign of inequality 

 in one or more of the criteria of imaginary roots is replaced by the sign 

 of equality is not adverted to, except in so far as to notice (6), that one 

 pair at least is then implied. There is another case, too, namely, that 

 in which consecutive zeros occur among the terms of equation, which 

 has not been specially considered above. The two cases have a certain 

 relation to each other, and it remains for us to examine what are the 

 inferences which these peculiarities justify. We shall first consider 

 the case of consecutive zeros. 



(21) — 1. When there are consecutive zeros. If all the terms between 

 the first term and the last are zeros, as in the equation 



the exact number of imaginary roots is determinable at once ; for since 

 then 



it is obvious that if n be even, and A 0 negative, there will be just two 

 real roots, numerically equal, but of opposite signs ; and, therefore, 

 n - 2 imaginary roots; while if A 0 be positive, all the roots will be 

 imaginary. 



But if n be odd, then, whether A 0 be negative or positive, there 

 will be only one real root ; and, therefore, n - 1 imaginary roots. 



The terms of an equation of degree n, being n + 1 in number, the 

 foregoing zeros are n - 1 in number. Hence, if we apply the rule at 

 page 355 to the equation [1] above, for the purpose of marking the number 

 of imaginary pairs, we must evidently write minus under the first zero; 

 and then, to secure conformity with the foregoing conclusions, must 

 write the signs plus and minus alternately, till the last zero is reached, 

 the sign under which, if n be even, must always be the opposite of the 

 sign of A 0 ; but if n be odd, this sign may be + or - indifferently. 



A n x n 



+ 0 + 0 + 0+ .. .. + 0 +A 0 = 0 . . . [1] 



