222

NEW METHOD FOR

TRACT 11

Then trying 3 for x, it is found to succeed. And dividingx l — 7a 2 + 13a’ — 18 by .r — 3, the quotient is x z — 4.r +6=0,a quadratic equation whose roots are 2 + ^—2 and 2 — V — 2,the two other roots of the proposed equation, which are bothimpossible or imaginary.

11. If the equation be .r 3 — 6a’ 2 + I4.r — 12 — 0 ;we shall havep = 6, q — 14, and r — 12 ; then x =

6 3 —27x 12

9 X 6 2 —

1

216

324 108 . . . .

—-=-=2. \V Inch being:

1x 14 9 36-42 54 b

substituted for x, it is found to answer, the sum of the termscoming out = 0. Therefore the roots are in arithmeticalprogression. And, accordingly, by dividing a’ 3 — 6.a 2 + 14.r— 12 by x — 2, the quotient is x~ — 4.r + 6=0, the rootsof which quadratic equation are 2-1- +■— 2 and 2 — —2,

the two other roots of the proposed equation, and the com-mon difference of the three roots is s / — 2.

12. But if the equation be a' 3 — Sa” + 22a' — 24 = 0 ;we shall have p = 8, q = 22, and r = 24 ; then x =

1 S 3 —27x 24 1 512- 648 13S_68_ ^

S 2 - 3 x 22 ~9 * 64 — 66 — HT ~ 17 ~ ‘

Which being substituted for a’ in the proposed equation, thesum of the terms differs very widely from the truth, therebyshowing that the middle root of the equation is an imaginaryone, as it is indeed, the three roots being 4, and 2+ +" — 2,and 2 ■— \/ — 2.

13. In Art. 2 the value of x was determined by assumingthe second terms of the two equations, equal to each other.But a like near value might be determined by assuming citherthe two third terms, or the two fourth terms equal.

(a’ 3 — 3a.v 2 + 3 arx — a? — 0,Thus the equations being | ^ _ p pl + qx _ r _ 0>

if we assume the third terms 3 a\v and qx equal, or a—^/^q^the sums of the second and fourth terms will be equal, namely,3 ux r + a 3 = px + r ; and hence we find

, « 3 -?- ,( \/~ ( /Y — r

x-J -— = V-———j

p — 3a p — -Vt?

by substituting \/-)q the value of a instead of it,