108 A NEW SOLUTION 



i. Compute, M s 3 -f- 2A + B 

 N -A + 2B+3C 

 ot = BN — 3CM 

 « = 3N — AM, 

 And Q = (4B 1 + 27O) — 2AC ( 9 B — A a ) + A 1 (2AC — B a ) 

 And let it be carefully noted under which of the two follow- 

 ing cafes the equation comes : 



Case I. When QJs negative. 

 Case II. When CHs pofitive. 

 To thefe two cafes a third may be added, viz. when Qj= o : 

 but of this cafe I ihall treat in one of the following examples. 



2. Compute alfo, a zr. ~l n ^ 



j - + n an 



~ V +30^ ~~ «" 



, M.i -I- N£ 



and rzz — M ± N ' 



3. Then, if the equation comes under cafe I : Find the an- 

 gle <p, of which the tangent is 7, the radius being unity : take 



z =ztan^,2 =. tan (~ + 120 ), and z zz tan (- -J- 240°) : And 



the three roots of the equation will be found, by fubftituting 



thefe values of z in the formula x zz ^Tg" 



4. But if the equation comes under Cafe II. we muft com- 

 pute 



T 3/ 1— T 



z — , . j 



And the only root of the equation will be found, by fubftitu- 

 ting this value of z in the formula x — " b , ^ 



The 



