430 
ME.  A.  CAYLEY’S  EIETH  ME2IOIE  UPOX  QrAXTICS. 
(a,  b,  — B^)"U=4n, 
which  expresses  that  the  provectant  of  the  quadric  is  equal  to  the  discrimiuaiit ; 
(a,  h,  cJhx-\-cy,  —ax—hyy=nV, 
which  expresses  that  a transmutant  of  the  quadric  is  equal  to  the  product  of  the  qua- 
dric and  the  discriminant. 
94.  When  the  quadric  is  expressed  in  terms  of  the  roots,  we  have 
={x—uy)(x—(5y% 
= /3)"; 
and  in  the  case  of  a pair  of  equal  roots, 
a~^JJ  =(x~ayf, 
□ =0. 
95.  The  problem  of  the  solution  of  a quadratic  equation  is  that  of  finding  a linear 
factor  of  the  quadric.  To  obtain  such  linear  factor  in  a symmetrical  form,  it  is  neces- 
sary to  introduce  arbitrary  quantities  which  do  not  really  enter  into  the  solution,  and 
the  form  obtained  is  thus  in  some  sort  more  complicated  than  in  the  like  problem  for  a 
cubic  or  a quartic.  The  solution  depends  on  the  linear  transformation  of  the  quadric, 
riz.  if  we  write 
{a,  b,  cjpc^-(^y,  vx-[-^yf={a!,  b',  djx,  y)\ 
so  that 
a!={a,  b,  vf, 
b'={a,  b,  cj\,  §), 
c'=(a,  b,  f)^ 
then 
a!d — b'^={ac—b^){X§—(jjvy, 
an  equation  which  in  a different  notation  is 
(a,  b,  cjx,  y)\(a,  b,  cJX,  ¥)>-(«,  b,  cl^:,  yJX,  Y)1>=  D(Ya,--Xy)*, 
in  which  form  it  is  a theorem  relating  to  the  quadric  and  its  first  and  second  emanants. 
The  equation  shows  that 
{«,  5,  cjs,  yJX,  Y)+^^(Yx-Xy), 
where  (X,  Y)  are  treated  as  supernumerary  arbitrary  constants,  is  a linear  factor  of 
{a,  b,  cjje,  yY,  and  this  is  the  required  solution. 
96.  In  the  case  of  two  quadrics,  the  expressions  considered  are 
{a,  b,  cXx,  y)\  (1) 
(a',  5',  djx,  y)\  (2) 
ac—b^  , (3) 
ad-2bV-^ca!  , (4) 
a!d-d^  , (5) 
