How do you simplify and find the restrictions for (x^2+3x-18)/(x^2-36)?

Nov 16, 2017

The restricted values of $x$ are: ${x}_{1} = 6$ and ${x}_{2} = - 6$.

Explanation:

There is no simplification possible.

To find the restrictions you have to see for what values of $x$ there's no solution. That would be when the denominatior is $0$.

So you get,
${x}^{2} - 36 = 0$

you isolate $x$,
${x}^{2} = 36$

and you do the square root in both sides,
$x = \pm \sqrt{36} = \pm 6$

so these are the restricted values on x:
${x}_{1} = 6$
${x}_{2} = - 6$

Nov 16, 2017

$f \left(x\right) =$$\frac{{x}^{2} + 3 x - 18}{{x}^{2} - 36} = \frac{\left(x - 3\right) \left(x + 6\right)}{\left(x - 6\right) \left(x + 6\right)} = \frac{x - 3}{x - 6}$

( $f \left(x\right) = 0$$\iff$$x = 3$

• x≠6 and x≠-6
$D f = \left(- \infty , - 6\right) U \left(- 6 , 6\right) U \left(6 , + \infty\right)$
$D f = R - \left\{- 6 , 6\right\}$ )
Nov 16, 2017

$\frac{{x}^{2} + 3 x - 18}{{x}^{2} - 36}$ simplifies to $\frac{x - 3}{x - 6}$ with the restriction that $x \ne 6$ and $x \ne + 6$

Explanation:

$\frac{{x}^{2} + 3 x - 18}{{x}^{2} - 36}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{\left(x + 6\right) \left(x - 3\right)}{\left(x + 6\right) \left(x - 6\right)}$

$\textcolor{w h i t e}{\text{XXX}}$Note the division is only defined if $x \ne \pm 6$

$\textcolor{w h i t e}{\text{XXX}} = \frac{x - 3}{x - 6}$ provided $\left(x + 6\right) \ne 0$