Solve the Inequality for x (x^2-3x-18)/(13x-x^2-42)>=0

$\frac{x^{2} - 3 x - 18}{13 x - x^{2} - 42} \geq 0$

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

$x^{2} - 3 x - 18 = 0$

$13 x - x^{2} - 42 = 0$

Factor $x^{2} - 3 x - 18$ 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 $- 18$ and whose sum is $- 3$ .

$- 6, 3$

Write the factored form using these integers.

$(x - 6) (x + 3) = 0$

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

$x - 6 = 0$

$x + 3 = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x - 6) (x + 3) = 0$ true.

$x = 6, - 3$

Factor the left side of the equation.

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$13 u - u^{2} - 42 = 0$

Factor by grouping.

Reorder terms.

$- u^{2} + 13 u - 42 = 0$

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 42 = 42$ and whose sum is $b = 13$ .

Factor $13$ out of $13 u$ .

$- u^{2} + 13 (u) - 42 = 0$

Rewrite $13$ as $6$ plus $7$

$- u^{2} + (6 + 7) u - 42 = 0$

Apply the distributive property.

$- u^{2} + 6 u + 7 u - 42 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(- u^{2} + 6 u) + 7 u - 42 = 0$

Factor out the greatest common factor (GCF) from each group.

$u (- u + 6) - 7 (- u + 6) = 0$

Factor the polynomial by factoring out the greatest common factor, $- u + 6$ .

$(- u + 6) (u - 7) = 0$

Replace all occurrences of $u$ with $x$ .

$(- x + 6) (x - 7) = 0$

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

$- x + 6 = 0$

$x - 7 = 0$

Set $- x + 6$ equal to $0$ and solve for $x$ .

Set $- x + 6$ equal to $0$ .

$- x + 6 = 0$

Solve $- x + 6 = 0$ for $x$ .

Subtract $6$ from both sides of the equation.

$- x = - 6$

Divide each term in $- x = - 6$ by $- 1$ and simplify.

Divide each term in $- x = - 6$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 6}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 6}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 6}{- 1}$

Simplify the right side.

Divide $- 6$ by $- 1$ .

$x = 6$

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

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

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

The final solution is all the values that make $(- x + 6) (x - 7) = 0$ true.

$x = 6, 7$

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

$x = 6, - 3$

$x = 6, 7$

Consolidate the solutions.

$x = 6, - 3, 6, 7$

Find the domain of $\frac{x^{2} - 3 x - 18}{13 x - x^{2} - 42}$ .

Set the denominator in $\frac{x^{2} - 3 x - 18}{13 x - x^{2} - 42}$ equal to $0$ to find where the expression is undefined.

$13 x - x^{2} - 42 = 0$

Solve for $x$ .

Factor the left side of the equation.

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$13 u - u^{2} - 42 = 0$

Factor by grouping.

Reorder terms.

$- u^{2} + 13 u - 42 = 0$

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 42 = 42$ and whose sum is $b = 13$ .

Factor $13$ out of $13 u$ .

$- u^{2} + 13 (u) - 42 = 0$

Rewrite $13$ as $6$ plus $7$

$- u^{2} + (6 + 7) u - 42 = 0$

Apply the distributive property.

$- u^{2} + 6 u + 7 u - 42 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(- u^{2} + 6 u) + 7 u - 42 = 0$

Factor out the greatest common factor (GCF) from each group.

$u (- u + 6) - 7 (- u + 6) = 0$

Factor the polynomial by factoring out the greatest common factor, $- u + 6$ .

$(- u + 6) (u - 7) = 0$

Replace all occurrences of $u$ with $x$ .

$(- x + 6) (x - 7) = 0$

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

$- x + 6 = 0$

$x - 7 = 0$

Set $- x + 6$ equal to $0$ and solve for $x$ .

Set $- x + 6$ equal to $0$ .

$- x + 6 = 0$

Solve $- x + 6 = 0$ for $x$ .

Subtract $6$ from both sides of the equation.

$- x = - 6$

Divide each term in $- x = - 6$ by $- 1$ and simplify.

Divide each term in $- x = - 6$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 6}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 6}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 6}{- 1}$

Simplify the right side.

Divide $- 6$ by $- 1$ .

$x = 6$

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

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

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

The final solution is all the values that make $(- x + 6) (x - 7) = 0$ true.

$x = 6, 7$

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

$(- \infty, 6) \cup (6, 7) \cup (7, \infty)$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 6$

$6 < x < 7$

$x > 7$

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 $x < - 3$ to see if it makes the inequality true.

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

$x = - 6$

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

$\frac{{(- 6)}^{2} - 3 \cdot - 6 - 18}{13 (- 6) - {(- 6)}^{2} - 42} \geq 0$

The left side $- 0.\overline{230769}$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 3 < x < 6$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 6$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{{(0)}^{2} - 3 \cdot 0 - 18}{13 (0) - {(0)}^{2} - 42} \geq 0$

The left side $0.\overline{428571}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $6 < x < 7$ to see if it makes the inequality true.

Choose a value on the interval $6 < x < 7$ and see if this value makes the original inequality true.

$x = 6.5$

Replace $x$ with $6.5$ in the original inequality.

$\frac{{(6.5)}^{2} - 3 \cdot 6.5 - 18}{13 (6.5) - {(6.5)}^{2} - 42} \geq 0$

The left side $19$ is greater than the right side $0$ , which means that the given statement is always true.

True

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.

$\frac{{(10)}^{2} - 3 \cdot 10 - 18}{13 (10) - {(10)}^{2} - 42} \geq 0$

The left side $- 4.\overline{3}$ is less 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.

$x < - 3$ False

$- 3 < x < 6$ True

$6 < x < 7$ True

$x > 7$ False

$x < - 3$ False

$- 3 < x < 6$ True

$6 < x < 7$ True

$x > 7$ False

The solution consists of all of the true intervals.

$- 3 \leq x < 6$ or $6 < x < 7$

The result can be shown in multiple forms.

Inequality Form:

$- 3 \leq x < 6 or 6 < x < 7$

Interval Notation:

$[- 3, 6) \cup (6, 7)$

Solve the Inequality for x (x^2-3x-18)/(13x-x^2-42)>=0

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