Solve the Inequality for x 4x^2-12>=-4x-9

$4 x^{2} - 12 \geq - 4 x - 9$

Step 1

Add $4 x$ to both sides of the inequality.

$4 x^{2} - 12 + 4 x \geq - 9$

Step 2

Convert the inequality to an equation.

$4 x^{2} - 12 + 4 x = - 9$

Step 3

Add $9$ to both sides of the equation.

$4 x^{2} - 12 + 4 x + 9 = 0$

Step 4

Add $- 12$ and $9$ .

$4 x^{2} + 4 x - 3 = 0$

Step 5

Factor by grouping.

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 $a \cdot c = 4 \cdot - 3 = - 12$ and whose sum is $b = 4$ .

Factor $4$ out of $4 x$ .

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

Rewrite $4$ as $- 2$ plus $6$

$4 x^{2} + (- 2 + 6) x - 3 = 0$

Apply the distributive property.

$4 x^{2} - 2 x + 6 x - 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(4 x^{2} - 2 x) + 6 x - 3 = 0$

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

$2 x (2 x - 1) + 3 (2 x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(2 x - 1) (2 x + 3) = 0$

Step 6

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

$2 x - 1 = 0$

$2 x + 3 = 0$

Step 7

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Step 8

Set $2 x + 3$ equal to $0$ and solve for $x$ .

Set $2 x + 3$ equal to $0$ .

$2 x + 3 = 0$

Solve $2 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Divide each term in $2 x = - 3$ by $2$ and simplify.

Divide each term in $2 x = - 3$ by $2$ .

$\frac{2 x}{2} = \frac{- 3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 3}{2}$

Divide $x$ by $1$ .

$x = \frac{- 3}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{2}$

Step 9

The final solution is all the values that make $(2 x - 1) (2 x + 3) = 0$ true.

$x = \frac{1}{2}, - \frac{3}{2}$

Step 10

Use each root to create test intervals.

$x < - \frac{3}{2}$

$- \frac{3}{2} < x < \frac{1}{2}$

$x > \frac{1}{2}$

Step 11

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 $x < - \frac{3}{2}$ to see if it makes the inequality true.

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

$x = - 4$

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

$4 {(- 4)}^{2} - 12 \geq - 4 \cdot - 4 - 9$

The left side $52$ is greater than the right side $7$ , which means that the given statement is always true.

True

Test a value on the interval $- \frac{3}{2} < x < \frac{1}{2}$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{3}{2} < x < \frac{1}{2}$ and see if this value makes the original inequality true.

$x = 0$

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

$4 {(0)}^{2} - 12 \geq - 4 \cdot 0 - 9$

The left side $- 12$ is less than the right side $- 9$ , which means that the given statement is false.

False

Test a value on the interval $x > \frac{1}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{1}{2}$ and see if this value makes the original inequality true.

$x = 3$

Replace $x$ with $3$ in the original inequality.

$4 {(3)}^{2} - 12 \geq - 4 \cdot 3 - 9$

The left side $24$ is greater than the right side $- 21$ , which means that the given statement is always true.

True

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

$x < - \frac{3}{2}$ True

$- \frac{3}{2} < x < \frac{1}{2}$ False

$x > \frac{1}{2}$ True

$x < - \frac{3}{2}$ True

$- \frac{3}{2} < x < \frac{1}{2}$ False

$x > \frac{1}{2}$ True

Step 12

The solution consists of all of the true intervals.

$x \leq - \frac{3}{2}$ or $x \geq \frac{1}{2}$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$x \leq - \frac{3}{2} or x \geq \frac{1}{2}$

Interval Notation:

$(- \infty, - \frac{3}{2}] \cup [\frac{1}{2}, \infty)$

Step 14

Solve the Inequality for x 4x^2-12>=-4x-9

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