Solve the Inequality for x 6x^2-x-6<=4x-2

$6 x^{2} - x - 6 \leq 4 x - 2$

Step 1

Move all terms containing $x$ to the left side of the inequality.

Subtract $4 x$ from both sides of the inequality.

$6 x^{2} - x - 6 - 4 x \leq - 2$

Subtract $4 x$ from $- x$ .

$6 x^{2} - 5 x - 6 \leq - 2$

Step 2

Convert the inequality to an equation.

$6 x^{2} - 5 x - 6 = - 2$

Step 3

Add $2$ to both sides of the equation.

$6 x^{2} - 5 x - 6 + 2 = 0$

Step 4

Add $- 6$ and $2$ .

$6 x^{2} - 5 x - 4 = 0$

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 $a \cdot c = 6 \cdot - 4 = - 24$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 x$ .

$6 x^{2} - 5 x - 4 = 0$

Rewrite $- 5$ as $3$ plus $- 8$

$6 x^{2} + (3 - 8) x - 4 = 0$

Apply the distributive property.

$6 x^{2} + 3 x - 8 x - 4 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(6 x^{2} + 3 x) - 8 x - 4 = 0$

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

$3 x (2 x + 1) - 4 (2 x + 1) = 0$

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

$(2 x + 1) (3 x - 4) = 0$

Step 6

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

$2 x + 1 = 0$

$3 x - 4 = 0$

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 $x$ .

Subtract $1$ from 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 $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 8

Set $3 x - 4$ equal to $0$ and solve for $x$ .

Set $3 x - 4$ equal to $0$ .

$3 x - 4 = 0$

Solve $3 x - 4 = 0$ for $x$ .

Add $4$ to both sides of the equation.

$3 x = 4$

Divide each term in $3 x = 4$ by $3$ and simplify.

Divide each term in $3 x = 4$ by $3$ .

$\frac{3 x}{3} = \frac{4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{4}{3}$

Divide $x$ by $1$ .

$x = \frac{4}{3}$

Step 9

The final solution is all the values that make $(2 x + 1) (3 x - 4) = 0$ true.

$x = - \frac{1}{2}, \frac{4}{3}$

Step 10

Use each root to create test intervals.

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

$- \frac{1}{2} < x < \frac{4}{3}$

$x > \frac{4}{3}$

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

$6 {(- 3)}^{2} - (- 3) - 6 \leq 4 (- 3) - 2$

The left side $51$ is greater than the right side $- 14$ , which means that the given statement is false.

False

Test a value on the interval $- \frac{1}{2} < x < \frac{4}{3}$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{1}{2} < x < \frac{4}{3}$ and see if this value makes the original inequality true.

$x = 0$

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

$6 {(0)}^{2} - (0) - 6 \leq 4 (0) - 2$

The left side $- 6$ is less than the right side $- 2$ , which means that the given statement is always true.

True

Test a value on the interval $x > \frac{4}{3}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{4}{3}$ and see if this value makes the original inequality true.

$x = 4$

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

$6 {(4)}^{2} - (4) - 6 \leq 4 (4) - 2$

The left side $86$ is greater than the right side $14$ , which means that the given statement is false.

False

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

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

$- \frac{1}{2} < x < \frac{4}{3}$ True

$x > \frac{4}{3}$ False

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

$- \frac{1}{2} < x < \frac{4}{3}$ True

$x > \frac{4}{3}$ False

Step 12

The solution consists of all of the true intervals.

$- \frac{1}{2} \leq x \leq \frac{4}{3}$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$- \frac{1}{2} \leq x \leq \frac{4}{3}$

Interval Notation:

$[- \frac{1}{2}, \frac{4}{3}]$

Step 14

Solve the Inequality for x 6x^2-x-6<=4x-2

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