Question

Solve for x:

Answer 1

Solution :

c)

$\frac{1}{x+2} - \frac{x}{2-x} - \frac{x+6}{x^2-4} = \frac{1}{x+2} + \frac{x}{x-2} - \frac{x+6}{x^2-4}$

$= \frac{(x-2) + x(x+2)}{x^2-4} - \frac{x+6}{x^2-4} = \frac{x^2 + 2x -8}{x^2-4}$

Thus, the given equation is reduced to :

$\frac{x^2 + 2x -8}{x^2-4} = 0$


一. + 2x-8=0

$\Rightarrow (x-2)(x+4) = 0$

Thus, {2,-4} is the set of roots of the last equation.

However, notice that the initial equation is not defined at x = 2.

Hence, the only root of the original equation is -4.

Note that 2 is not a root of the initial equation. It appears as a root (of the last equation) because of the increase in the degree of the equation. Such a root is called an Extraneous Root of the initial equation.
However, the only root of the original equation is - 4.