Solve for x (x^2-36)/(x-6)<=0

$\frac{x^{2} - 36}{x - 6} \leq 0$

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x^{2} - 36 = 0$

$x - 6 = 0$

Add $36$ to both sides of the equation.

$x^{2} = 36$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{36}$

Simplify $\pm \sqrt{36}$ .

Rewrite $36$ as $6^{2}$ .

$x = \pm \sqrt{6^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 6$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 6$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 6$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 6, - 6$

Add $6$ to both sides of the equation.

$x = 6$

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 6, - 6$

$x = 6$

Consolidate the solutions.

$x = 6, - 6, 6$

Find the domain of $\frac{x^{2} - 36}{x - 6}$ .

Set the denominator in $\frac{x^{2} - 36}{x - 6}$ equal to $0$ to find where the expression is undefined.

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

The domain is all values of $x$ that make the expression defined.

$(- \infty, 6) \cup (6, \infty)$

The solution consists of all of the true intervals.

$x \leq - 6$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 6$

Interval Notation:

$(- \infty, - 6]$