Solve for x (x^2+x-12)/(x^2-4x+4)>0

$\frac{x^{2} + x - 12}{x^{2} - 4 x + 4} > 0$

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

$x^{2} + x - 12 = 0$

$x^{2} - 4 x + 4 = 0$

Factor $x^{2} + x - 12$ 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 $- 12$ and whose sum is $1$ .

$- 3, 4$

Write the factored form using these integers.

$(x - 3) (x + 4) = 0$

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

$x - 3 = 0$

$x + 4 = 0$

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

Set $x - 3$ 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 $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

$x = 3, - 4$

Factor using the perfect square rule.

Rewrite $4$ as $2^{2}$ .

$x^{2} - 4 x + 2^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 2 + 2^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 2$ .

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

Set the $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

$x = 3, - 4$

$x = 2$

Consolidate the solutions.

$x = 3, - 4, 2$

Find the domain of $\frac{x^{2} + x - 12}{x^{2} - 4 x + 4}$ .

Set the denominator in $\frac{x^{2} + x - 12}{x^{2} - 4 x + 4}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 4 x + 4 = 0$

Solve for $x$ .

Factor using the perfect square rule.

Rewrite $4$ as $2^{2}$ .

$x^{2} - 4 x + 2^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 2 + 2^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 2$ .

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

Set the $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

$(- \infty, 2) \cup (2, \infty)$

Use each root to create test intervals.

$x < - 4$

$- 4 < x < 2$

$2 < x < 3$

$x > 3$

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 $x < - 4$ 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.

$\frac{{(- 6)}^{2} - 6 - 12}{{(- 6)}^{2} - 4 \cdot - 6 + 4} > 0$

The left side $0.28125$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 4 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $- 4 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{{(0)}^{2} + 0 - 12}{{(0)}^{2} - 4 \cdot 0 + 4} > 0$

The left side $- 3$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $2 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 3$ and see if this value makes the original inequality true.

$x = 2.5$

Replace $x$ with $2.5$ in the original inequality.

$\frac{{(2.5)}^{2} + 2.5 - 12}{{(2.5)}^{2} - 4 \cdot 2.5 + 4} > 0$

The left side $- 13$ is not greater than the right side $0$ , which means that the given statement is false.

False

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.

$\frac{{(6)}^{2} + 6 - 12}{{(6)}^{2} - 4 \cdot 6 + 4} > 0$

The left side $1.875$ 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.

$x < - 4$ True

$- 4 < x < 2$ False

$2 < x < 3$ False

$x > 3$ True

$x < - 4$ True

$- 4 < x < 2$ False

$2 < x < 3$ False

$x > 3$ True

The solution consists of all of the true intervals.

$x < - 4$ or $x > 3$

The result can be shown in multiple forms.

Inequality Form:

$x < - 4 or x > 3$

Interval Notation:

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

Solve for x (x^2+x-12)/(x^2-4x+4)>0

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