Solve the Inequality for x (2x+5)/(x^2+2x-35)>=0

$\frac{2 x + 5}{x^{2} + 2 x - 35} \geq 0$

Step 1

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

$2 x + 5 = 0$

$x^{2} + 2 x - 35 = 0$

Step 2

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Step 3

Divide each term in $2 x = - 5$ by $2$ and simplify.

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

Step 4

Factor $x^{2} + 2 x - 35$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 35$ and whose sum is $2$ .

$- 5, 7$

Write the factored form using these integers.

$(x - 5) (x + 7) = 0$

Step 5

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

$x - 5 = 0$

$x + 7 = 0$

Step 6

Set $x - 5$ equal to $0$ and solve for $x$ .

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Step 7

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 8

The final solution is all the values that make $(x - 5) (x + 7) = 0$ true.

$x = 5, - 7$

Step 9

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

$x = - \frac{5}{2}$

$x = 5, - 7$

Step 10

Consolidate the solutions.

$x = - \frac{5}{2}, 5, - 7$

Step 11

Find the domain of $\frac{2 x + 5}{x^{2} + 2 x - 35}$ .

Set the denominator in $\frac{2 x + 5}{x^{2} + 2 x - 35}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 2 x - 35 = 0$

Solve for $x$ .

Factor $x^{2} + 2 x - 35$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 35$ and whose sum is $2$ .

$- 5, 7$

Write the factored form using these integers.

$(x - 5) (x + 7) = 0$

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

$x - 5 = 0$

$x + 7 = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $(x - 5) (x + 7) = 0$ true.

$x = 5, - 7$

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

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

Step 12

Use each root to create test intervals.

$x < - 7$

$- 7 < x < - \frac{5}{2}$

$- \frac{5}{2} < x < 5$

$x > 5$

Step 13

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.

$\frac{2 (- 10) + 5}{{(- 10)}^{2} + 2 (- 10) - 35} \geq 0$

The left side $- 0.\overline{3}$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 7 < x < - \frac{5}{2}$ to see if it makes the inequality true.

Choose a value on the interval $- 7 < x < - \frac{5}{2}$ and see if this value makes the original inequality true.

$x = - 5$

Replace $x$ with $- 5$ in the original inequality.

$\frac{2 (- 5) + 5}{{(- 5)}^{2} + 2 (- 5) - 35} \geq 0$

The left side $0.25$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- \frac{5}{2} < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{5}{2} < x < 5$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{2 (0) + 5}{{(0)}^{2} + 2 (0) - 35} \geq 0$

The left side $- 0.\overline{142857}$ is less than the right side $0$ , which means that the given statement is false.

False

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.

$\frac{2 (8) + 5}{{(8)}^{2} + 2 (8) - 35} \geq 0$

The left side $0.4\overline{6}$ 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.

$x < - 7$ False

$- 7 < x < - \frac{5}{2}$ True

$- \frac{5}{2} < x < 5$ False

$x > 5$ True

$x < - 7$ False

$- 7 < x < - \frac{5}{2}$ True

$- \frac{5}{2} < x < 5$ False

$x > 5$ True

Step 14

The solution consists of all of the true intervals.

$- 7 < x \leq - \frac{5}{2}$ or $x > 5$

Step 15

The result can be shown in multiple forms.

Inequality Form:

$- 7 < x \leq - \frac{5}{2} or x > 5$

Interval Notation:

$(- 7, - \frac{5}{2}] \cup (5, \infty)$

Step 16

Solve the Inequality for x (2x+5)/(x^2+2x-35)>=0

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