# How do you find the domain of f(x)=(x-3)/(x^2-x-2)?

Nov 9, 2015

$x \in \mathbb{R}$ \ {-1; 2}

#### Explanation:

To find the domain of a rational function, you need to determine all values of $x$ for which the demoninator is equal to $0$.

So, set ${x}^{2} - x - 2 = 0$ and solve the quadratic equation.

There are several possibilities to solve a quadratic equation, one of my favourite is "completing the circle":

${x}^{2} - x - 2 = 0$
$\iff {x}^{2} - x = 2$

Try to write the left side of the equation like ${\left(x - a\right)}^{2} = {x}^{2} - 2 a x + {a}^{2}$.
We already have the ${x}^{2}$, and from $- x = - 2 a x$ we can conclude that $a = \frac{1}{2}$. So, the only term that is missing on the left side to complete the quadratic form is $+ \frac{1}{4}$.

In order to prevail the equality, the term needs to be added on both sides of the equation, leading us to the following:
${x}^{2} - x + \frac{1}{4} = 2 + \frac{1}{4}$
$\iff {\left(x - \frac{1}{2}\right)}^{2} = \frac{9}{4}$

Now, we can calculate the root on both sides. Don't forget that the root of $\frac{9}{4}$ has two solutions since both ${\left(\frac{3}{2}\right)}^{2} = \frac{9}{4}$ and ${\left(- \frac{3}{2}\right)}^{2} = 94$ hold.

Thus, we can further solve the quadratic equation as follows:
$x - \frac{1}{2} = \frac{3}{2} \mathmr{and} x - \frac{1}{2} = - \frac{3}{2}$
$\iff x = 2 \mathmr{and} x = - 1$

This means that for those two values, the denominator of our function would be $0$ which is not admissible.

So, the domain of the function are all real numbers except $x = 2$ and $x = - 1$.
$x \in \mathbb{R}$ \ {-1; 2}