xp2−(x−a)2=0
we have difference of squares
(xp−(x−a))(xp+(x−a))=0
Two equations and two solutions:
first equation:
xp−x+a=0
p=xx−a=x−xa
first solution:
z=32x23−2ax+C
second equation:
xp+x−a=0
p=−xx−a=−x+xa
second solution:
z=−32x23+2ax+C
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