For
,
should be a perfect square of = 0.
Equating
to 0, we get:
= 0
= 0


1or –5
Substituting y = 1 in the original equation, we get:
, but this cannot be true, as
, hence, 1 is not the correct solution to the equation.
Substituting y = –5 in the original equation, we get:
, this is true, as the equation is satisfied. Hence, –5 is the correct solution to the equation. Thus there is only a single solution to the equation.
Note that we are not checking by equating with perfect squares, as:
= 4 (a perfect square)
= 16


3 or –7
Substituting y = 3 in the original equation, we get:
, but this cannot be true, as
, hence, 3 is not the correct solution to the equation.
Substituting y = –7 in the original equation, we get:
, but this cannot be true, as
, hence, –7 is not the correct solution to the equation.