Problem: If ab=0 and ∣a∣=∣b∣ the number of distinct values of x satisfying the equationbx−a+ax−b=x−ab+x−ba, is:
Answer Choices:
A. zero
B. one
C. two
D. three
E. four
Solution:
a(x−a)2(x−b)+b(x−b)2(x−a)=ab2(x−b)+a2b(x−a)
a(x−a)[(x−a)(x−b)−ab]=−b(x−b)[(x−a)(x−b)−ab]
∴[(x−a)(x−b)−ab)[ax−a2+bx−b2]=0
x2−(a+b)x=0 or (a+b)x=a2+b2∴x=0,x=a+b, or x=a+ba2+b2