Imaginary Roots. Thus the number of unequal roots of the equation of the 9th degree, (x - 7)(2+4)3 (2 — 195 = 0, is but three, namely, 7,-4, and 1, and yet it is to be regarded as having 9 roots, one equal to 7, three equal to -4, and five equal to 1. 271. Corollary. The equation x = a would appear to have but one root, that is, a; but it must, by art. 269, have n roots, or rather, the nth root of a must have n different values. 272. EXAMPLES. 1. Find the two roots of the equation 22 = 1. Ans. 2=1, or = -1. 2. Find the three roots of the equation 23 = 1. Solution. Since one root of this equation is = 1, the equation 23-1=0 must be divisible by 2 — 1, and we have 23 – 1= (2x — 1) (22 ++1) = 0. Now the roots of the equation 22 +r+1=0 are Imaginary Roots. 24 = Hence the required roots are x=1,=i(-1+V-3), and=i(-1-V-3). 3. Find the four roots of the equation 1. Solution. The square root of this equation is = + 1, or +1, or = -1; so that the required roots are q=1,= -1,=V-1, and -V-1. 4. Find the five roots of the equation 25 = 1. Solution. , Since one root of this equation is = 1, the equation 25-1=0 must be divisible by 2 – . 1, and we have 25.—1=(x-1) (24 +23+2° +x+1)=0. Now the roots of the equation 24 +23+72++1=0 can be found by the following peculiar process. Divide by 22, and we have 22++1+1+ 0. If we make y=x+ Solution of Equations of a peculiar Form. which, being substituted in the preceding equation, gives y2 + y-1=0; the roots of which are y=i(-1+v5), and =i(-1-75). But the values of x deduced from the equation y=x+ or Y * = -1, are = =i[y tv (ye_4)], and=i[y-v(y2-4)], in which y, being substituted, gives X X x=\ [-1-752V (-10+2V5)], and =[-1-75+v(-10+2 v 5)]. 5. Find the six roots of the equation 26 = 1. Ans. q=1,=-1,=11-12V-3), and = (1+V-3). We might proceed in the same way to higher equations, such as the 8th, 9th, 12th, &c.; but, since much more simple solutions are given by the aid of trigonometry, this subject will be postponed to a more advanced part of the course. 273. Corollary: Before proceeding farther, we may remark, that the method of solution used in the last example of the preceding article may be applied equation of an even degree, in which the successive coefficients of the different powers of a are the same, whether the equation is arranged according to the ascending or according to the descend Solution of Equations of a peculiar Form. ing powers of x, as is the case in the following equation. Ackn + B @n-+ C 22"-2+ &c. + C 22 + B x +A=0. 274. EXAMPLES. 1. Solve the equation A 24 + B 23 + Cz? + Br+A=0, Solution. Divide by a?, and we have A (+*+1) +B(*+1)+C=0, y=++ and, if we make we have , and A yo +By+C-2A=0; the roots of which are 1 BIV (2 42-AC++ B2) A which are to be substituted in the values of x, x=iyv (ya — 1), deduced from the equation y=2+1 2. Solve the equation 26 +325 – 724 +6 23 _7*2 +31+1=0. Values of Coefficients in Equations. Solution. Divide by 203, and we have and if we make. (**+ **) +3(+2) –7(2+1)+6=0; y=+ + we have 1 = y2 — 2, 22 .5; 23 + 23 = y3 – 3 and the equation becomes, by substitution, y3 + 3 ya — 10 y = 0. The roots of this equation are y = 0, = 2, and =and, therefore, the values of x are x=V-1,= 1, or=i(-5+21). 3. Solve the equation 28 + 2 26 – 624 + 232 +1= 0. Ans. r=+1, or=uiv(-2+V3). 4. Solve the equation 2 x4 - 3x3 - 2-3x+2=0. 275. Corollary. It follows, from art. 269, that an equation of the second degree has two roots, both of which are given by the process of art. 230; and if the equation is reduced to the form 22 + ax + b = 0, and the roots denoted by x' and x', we have ." 22+ar+b=(x-) (-- ")=0. |