Question

Where is the hole in the equation (x-2)/(x^2-4)

Answer 1

Answer:

x = 2, or at the point (2, 1/4)

Step-by-step explanation:

Factor the denominator (difference of two squares).

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

There is a common factor of $x-2$ which indicates there is a hole at $x=2$ (where $x-2=0$).

If you're used to answering a question like this with the complete coordinates of a point, put $x=2$ into a simplified fraction, by cancelling the common factor.

Substitute 2 for x.

$\frac{1}{x+2} = \frac{1}{2+2} = \frac{1}{4}$

The hole is at the point (2, 1/4).