360 



5. 5x* - 2X 1 + 3x 6 + 0 + 0 + 0 + 4^ - 2x + 60 = 0 

 + - + - + -+- 



Therefore all the roots are imaginary. This conclusion may be verified 

 as follows. "Writing the equation thus : — 



x%bx % - 2x + 3) + (4a; 3 - 2x + 60) = 0, 



we see that each of the two quadratic expressions is positive whatever 

 real value be given to x ; and, therefore, as x Q is always positive too, 

 no real value of x can satisfy the equation. 



6. 3z 8 - 5x* + 0 - 2x? + 7a; 2 + 0 - 4 = 0 

 + + - + + + 



Consequently, only two imaginary roots are detected; they are in the 

 negative region. 



7* 



a 1 H0 + 0+0+0-^ 13 +0+0 + 0-&c 9 + e# 8 + 0 + dx*+0 + 0 + 0+ 0+0-5=0 

 + - + - + + - +- + + - + - +- + + 



Therefore, the equation has fourteen imaginary roots, at least ; there 

 may be sixteen, but there cannot be a greater number, since, as the sign 

 of the last term shows, two roots, at least, must be real. If this sign 

 had been +, then the sign under the last zero would have been - ; and 

 the equation would have had eight pairs of imaginary roots at least. 



(23) 2. When there are triads of equality . Let us first suppose that 

 all the triads, except the last triad, furnish conditions of equality. "We 

 except the last triad, because if every triad throughout were to satisfy 

 the condition of equality, the roots would all be real and equal ; that is 

 to 6ay, the equation would be of the form 



A n (x + r ) n - 0 ; 



so that 



would express each of the n roots. But if, as here supposed, the last 

 term A 0 of the equation be of such value as to render the final triad 

 one of inequality — all the preceding being triads of equality — then it is 

 plain that the form of the equation will be 



A n (x + r) n ^ c = 0 ; 



or 



{x + r)* = - -— 



* From Fourier—" Analyse des Equations D&ermin&s." 



