Solve the Inequality for x (x+5)/((x-3)^2(x-4))<=0

$\frac{x + 5}{{(x - 3)}^{2} (x - 4)} \leq 0$

Step 1

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

$x + 5 = 0$

${(x - 3)}^{2} = 0$

$x - 4 = 0$

Step 2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 3

Set the $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 4

Add $3$ to both sides of the equation.

$x = 3$

Step 5

Add $4$ to both sides of the equation.

$x = 4$

Step 6

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

$x = - 5$

$x = 3$

$x = 4$

Step 7

Consolidate the solutions.

$x = - 5, 3, 4$

Step 8

Find the domain of $\frac{x + 5}{{(x - 3)}^{2} (x - 4)}$ .

Set the denominator in $\frac{x + 5}{{(x - 3)}^{2} (x - 4)}$ equal to $0$ to find where the expression is undefined.

${(x - 3)}^{2} (x - 4) = 0$

Solve for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

${(x - 3)}^{2} = 0$

$x - 4 = 0$

Set ${(x - 3)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 3)}^{2}$ equal to $0$ .

${(x - 3)}^{2} = 0$

Solve ${(x - 3)}^{2} = 0$ for $x$ .

Set the $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make ${(x - 3)}^{2} (x - 4) = 0$ true.

$x = 3, 4$

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

$(- \infty, 3) \cup (3, 4) \cup (4, \infty)$

Step 9

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 3$

$3 < x < 4$

$x > 4$

Step 10

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Replace $x$ with $- 8$ in the original inequality.

$\frac{(- 8) + 5}{{((- 8) - 3)}^{2} ((- 8) - 4)} \leq 0$

The left side $0.00206611$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 5 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$\frac{(0) + 5}{{((0) - 3)}^{2} ((0) - 4)} \leq 0$

The left side $- 0.13\overline{8}$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $3 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $3 < x < 4$ and see if this value makes the original inequality true.

$x = 3.5$

Replace $x$ with $3.5$ in the original inequality.

$\frac{(3.5) + 5}{{((3.5) - 3)}^{2} ((3.5) - 4)} \leq 0$

The left side $- 68$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 4$ to see if it makes the inequality true.

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$\frac{(6) + 5}{{((6) - 3)}^{2} ((6) - 4)} \leq 0$

The left side $0.6\overline{1}$ is greater than the right side $0$ , which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 5$ False

$- 5 < x < 3$ True

$3 < x < 4$ True

$x > 4$ False

$x < - 5$ False

$- 5 < x < 3$ True

$3 < x < 4$ True

$x > 4$ False

Step 11

The solution consists of all of the true intervals.

$- 5 \leq x < 3$ or $3 < x < 4$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$- 5 \leq x < 3 or 3 < x < 4$

Interval Notation:

$[- 5, 3) \cup (3, 4)$

Step 13

Solve the Inequality for x (x+5)/((x-3)^2(x-4))<=0

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