Problem: If abξ =0 and β£aβ£ξ =β£bβ£ the number of distinct values of x satisfying the equationbxβaβ+axβbβ=xβabβ+xβbaβ, is:
Answer Choices:
A. zero
B. one
C. two
D. three
E. four
Solution:
a(xβa)2(xβb)+b(xβb)2(xβa)=ab2(xβb)+a2b(xβa)
a(xβa)[(xβa)(xβb)βab]=βb(xβb)[(xβa)(xβb)βab]
β΄[(xβa)(xβb)βab)[axβa2+bxβb2]=0
x2β(a+b)x=0 or (a+b)x=a2+b2β΄x=0,x=a+b, or x=a+ba2+b2β