1953 AHSME Problems/Problem 23
The equation has:
an extraneous root between
and
an extraneous root between
and
a true root between
and
two true roots
two extraneous roots
We multiply both sides by to get the equation
. We square both sides to get
, or
. We can factor the quadratic as
, giving us roots of
and
. We plug these values in to find that
is an extraneous root and that
is a true root, giving an answer of
.