Question

The number of real solutions of the equation $\frac{6-x}{x^2-4}=2+\frac{x}{x+2}$ is

• A. two
• B. one
• C. zero
• D. none of these

Answer 1

Answer: (B)

one

$\frac{6-x}{x^2-4}=2+\frac{x}{x+2}$
$\frac{6-x}{(x-2)(x+2)}=\frac{3x+4}{x+2}$$x\ne -2$
$\therefore 3x^2-x-14=0$.
The roots of this equation are $-2$ and $\frac{7}{3}$ but $-2$ is not possible
$\therefore x=\frac{7}{3}$.