A New Method of . certain Equations. 87 
Put o=r-+a, 
a a r+b, PHS $8024 al- 
«=r 4 Bar? +3br-+c ac=b’, 
mz? -+-3a’mz? +35 enter et 
2°+-3a'mz? + 3b’m mz+-e'm= 
: [1 —mM.z°. 
_ Assume (a'm)? =b'm 
_ Then also (a'm)*=c'm; For, (a'm)* _ (b’m)? =v m=e is 
a’m am 
Substitute values; 
23 +4-3(a'm).z?+-3(a'm)?.z+-(a’m)®?=1— m.z°. soz-a'm 
[=*Vl—mz 
az+a!m= 24/(a'® —a'®m.z, rcs sale Al a’? — B)a’.z. 
{—q!?)\q' z=—b, z= ——— 
(a +2/(b'—a’ (6 —a!?)a’.z= z= rae VO 
it Tear = 
omnes ~ rbat3/(b— (b—a*)(r-+-a) " 
From this we derive the 
Rute. 
To find x, when 2° -+-3axn*+3bhx+c=0 ; =0 5 
(A) Find ,inthe quadratic, b—a3.r? +c—ab.r--ac-b4 =o, 
r3 + 2ar+6 
(B) Put z ig eS Sasa 
rate (b —a?)(r-ta) 
And =z =z+r. 
In those questions which are embraced by the irreduci- 
ble case of Cardan, it will be seen that the value of r is 
imaginary. For then, if 7° +3b2+c=o0, br?+cr—b?=0, 
whence 2br+c=4/c?+46'; which gives an imaginary 
value, whenever 5° is negative and exceeds — When 
these two quantities become equal, the sidaeiaty part of 
the expression vanishes ; and, in the general solution, we 
have 26—a?.rt+c—ab=o.  (C) 
