Solve the Inequality for x 3x^2-1>=-2x

$3 x^{2} - 1 \geq - 2 x$

Step 1

Add $2 x$ to both sides of the inequality.

$3 x^{2} - 1 + 2 x \geq 0$

Step 2

Convert the inequality to an equation.

$3 x^{2} - 1 + 2 x = 0$

Step 3

Factor by grouping.

Reorder terms.

$3 x^{2} + 2 x - 1 = 0$

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 $a \cdot c = 3 \cdot - 1 = - 3$ and whose sum is $b = 2$ .

Factor $2$ out of $2 x$ .

$3 x^{2} + 2 (x) - 1 = 0$

Rewrite $2$ as $- 1$ plus $3$

$3 x^{2} + (- 1 + 3) x - 1 = 0$

Apply the distributive property.

$3 x^{2} - 1 x + 3 x - 1 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{2} - 1 x) + 3 x - 1 = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

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

Step 4

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

$3 x - 1 = 0$

$x + 1 = 0$

Step 5

Set $3 x - 1$ equal to $0$ and solve for $x$ .

Set $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Solve $3 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$3 x = 1$

Divide each term in $3 x = 1$ by $3$ and simplify.

Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Divide $x$ by $1$ .

$x = \frac{1}{3}$

Step 6

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 7

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

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

Step 8

Use each root to create test intervals.

$x < - 1$

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

$x > \frac{1}{3}$

Step 9

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 $x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

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

$3 {(- 4)}^{2} - 1 \geq - 2 \cdot - 4$

The left side $47$ is greater than the right side $8$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

$3 {(0)}^{2} - 1 \geq - 2 \cdot 0$

The left side $- 1$ is less than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 3$

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

$3 {(3)}^{2} - 1 \geq - 2 \cdot 3$

The left side $26$ is greater than the right side $- 6$ , which means that the given statement is always true.

True

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

$x < - 1$ True

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

$x > \frac{1}{3}$ True

$x < - 1$ True

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

$x > \frac{1}{3}$ True

Step 10

The solution consists of all of the true intervals.

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

Step 11

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

Step 12

Solve the Inequality for x 3x^2-1>=-2x

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