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 .