JAMB 1984 · UME · Q30

If $a = \dfrac{2x}{1 - x}$ and $b = \dfrac{1 + x}{1 - x}$ , then $a^2 - b^2$ in its simplest form is

Worked solution (try it first)

Both fractions have bottom $1 - x$ , so $a^2 - b^2 = \dfrac{4x^2 - (1 + x)^2}{(1 - x)^2}$ .
Expand the top: $4x^2 - (1 + 2x + x^2) = 3x^2 - 2x - 1$ , which factorises as $(3x + 1)(x - 1)$ .
The bottom $(1 - x)^2$ equals $(x - 1)^2$ , so one factor $x - 1$ cancels.
So $a^2 - b^2 = \dfrac{3x + 1}{x - 1}$ , option A.