646 PHILOSOPHICAL TRANSACTIONS. [aNNO 16Q4. 



tuting a + e for z, as before said, it will he±b=±se + tee. For the 

 better understanding of which, take the following examples. 



Example 1. Let the equation proposed bez*— 3zz + 75z= 10000. For 

 the first supposition take a = 10 ; hence will arise the following equation : 



+ 450 — 4015 e + 597 e e — 40 e^ + e* = 0. 



s t u 



The signs + and — , of e and e^, are left doubtful, till it be known whether 

 e is negative or affirmative, which may admit of some difficulty, since in equa- 

 tions that have many roots, the homogeneum comparationis, as it is called, 

 may be increased by diminishing the quantity a, and contrariwise may be 

 diminished when that is increased. But the sign of e is determined by the 

 sign of the quantity b : for the resolvend being taken away from the homoge- 

 neum formed of a, the sign of s e, and therefore of the parts prevailing in its 

 composition, will always be contrary to the sign of the difference b : hence it 

 will appear whether it should be — e or + e, or whether a be assumed greater 



or less than the true root. Now the quantity e is always =: 3-^-^^^^ — iJJJZ — ^ 

 when b and t have the same sign ; but when these have contrary signs, then is s = 

 — i^-— - — ^^^-^. Now if it be found that it must be — e, then in the affir- 

 mative parts of the equation let e, e^, c^ &c. be made negative, and in the 

 negative parts let them be made affirmative, that is, let them be all written 

 with the contrary sign. On the other hand, if it be -\- e, let those parts re- 

 tain their own signs. 



Now in our example we have 104 50 instead of the resolvend 10000, or 6 = 

 4- 450, whence it appears that a is assumed greater than just, and therefore it 

 must be — e. Hence the equation becomes 10450 — 4015 e -f- 597 ee — 40 e* 

 + e" = 10000 ; that is, 450 — 4015 e •{■ 5Q7 ee = O. Therefore 450 = 



4015 e — 597 e^j or b = s e — t e e, the root of which is e = "^-^-^ — ±fj_^ — ^ 



