Solve the Inequality for x 3x^2+x<2

$3 x^{2} + x < 2$

Convert the inequality to an equation.

$3 x^{2} + x = 2$

Subtract $2$ from both sides of the equation.

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

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 $a \cdot c = 3 \cdot - 2 = - 6$ and whose sum is $b = 1$ .

Multiply by $1$ .

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

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

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $3 x - 2$ .

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

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

$3 x - 2 = 0$

$x + 1 = 0$

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

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

$3 x - 2 = 0$

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

Add $2$ to both sides of the equation.

$3 x = 2$

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

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

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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.

$x = - 1$

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

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

Use each root to create test intervals.

$x < - 1$

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

$x > \frac{2}{3}$

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 $x < - 1$ 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.

$3 {(- 4)}^{2} - 4 < 2$

The left side $44$ is not less than the right side $2$ , which means that the given statement is false.

False

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

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

$x = 0$

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

$3 {(0)}^{2} + 0 < 2$

The left side $0$ is less than the right side $2$ , which means that the given statement is always true.

True

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

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

$x = 3$

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

$3 {(3)}^{2} + 3 < 2$

The left side $30$ is not less than the right side $2$ , which means that the given statement is false.

False

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

$x < - 1$ False

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

$x > \frac{2}{3}$ False

$x < - 1$ False

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

$x > \frac{2}{3}$ False

The solution consists of all of the true intervals.

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

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

$(- 1, \frac{2}{3})$

Solve the Inequality for x 3x^2+x<2

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