Solve the Inequality for x ((x+7)(x-3))/((x-5)^2)>0

$\frac{(x + 7) (x - 3)}{{(x - 5)}^{2}} > 0$

Step 1

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

$x + 7 = 0$

$x - 3 = 0$

${(x - 5)}^{2} = 0$

Step 2

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 3

Add $3$ to both sides of the equation.

$x = 3$

Step 4

Set the $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 5

Add $5$ to both sides of the equation.

$x = 5$

Step 6

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

$x = - 7$

$x = 3$

$x = 5$

Step 7

Consolidate the solutions.

$x = - 7, 3, 5$

Step 8

Find the domain of $\frac{(x + 7) (x - 3)}{{(x - 5)}^{2}}$ .

Set the denominator in $\frac{(x + 7) (x - 3)}{{(x - 5)}^{2}}$ equal to $0$ to find where the expression is undefined.

${(x - 5)}^{2} = 0$

Solve for $x$ .

Set the $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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

$(- \infty, 5) \cup (5, \infty)$

Step 9

Use each root to create test intervals.

$x < - 7$

$- 7 < x < 3$

$3 < x < 5$

$x > 5$

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 $x < - 7$ to see if it makes the inequality true.

Choose a value on the interval $x < - 7$ and see if this value makes the original inequality true.

$x = - 10$

Replace $x$ with $- 10$ in the original inequality.

$\frac{((- 10) + 7) ((- 10) - 3)}{{((- 10) - 5)}^{2}} > 0$

The left side $0.17\overline{3}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 7 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 7 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{((0) + 7) ((0) - 3)}{{((0) - 5)}^{2}} > 0$

The left side $- 0.84$ is not greater than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 4$

Replace $x$ with $4$ in the original inequality.

$\frac{((4) + 7) ((4) - 3)}{{((4) - 5)}^{2}} > 0$

The left side $11$ is greater than the right side $0$ , which means that the given statement is always true.

True

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.

$\frac{((8) + 7) ((8) - 3)}{{((8) - 5)}^{2}} > 0$

The left side $8.\overline{3}$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < - 7$ True

$- 7 < x < 3$ False

$3 < x < 5$ True

$x > 5$ True

$x < - 7$ True

$- 7 < x < 3$ False

$3 < x < 5$ True

$x > 5$ True

Step 11

The solution consists of all of the true intervals.

$x < - 7$ or $3 < x < 5$ or $x > 5$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$x < - 7 or 3 < x < 5 or x > 5$

Interval Notation:

$(- \infty, - 7) \cup (3, 5) \cup (5, \infty)$

Step 13

Solve the Inequality for x ((x+7)(x-3))/((x-5)^2)>0

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