Question

How many extraneous solutions does the equation below have?
StartFraction 2 m Over 2 m + 3 EndFraction minus StartFraction 2 m Over 2 m minus 3 EndFraction = 1
0
1
2
3

Answer 1

First restrict domain: 2m+3 ≠ 0 and 2m−3 ≠ 0, so m ≠ ±3/2.

Multiply both sides by (2m+3)(2m−3):
2m(2m−3) − 2m(2m+3) = (2m+3)(2m−3)
4m^2 − 6m − (4m^2 + 6m) = 4m^2 − 9
−12m = 4m^2 − 9
4m^2 + 12m − 9 = 0

Solve: m = [−12 ± √288]/8 = (−3 ± 3√2)/2, neither of which is ±3/2, so both are valid solutions and no excluded values were introduced.

Number of extraneous solutions = 0.