Solve the Inequality for x 2x^2+3x-2>0

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

Convert the inequality to an equation.

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

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 $a \cdot c = 2 \cdot - 2 = - 4$ and whose sum is $b = 3$ .

Factor $3$ out of $3 x$ .

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

Rewrite $3$ as $- 1$ plus $4$

$2 x^{2} + (- 1 + 4) x - 2 = 0$

Apply the distributive property.

$2 x^{2} - 1 x + 4 x - 2 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{2} - 1 x) + 4 x - 2 = 0$

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

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

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

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

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

$2 x - 1 = 0$

$x + 2 = 0$

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 $1$ .

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

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

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

Use each root to create test intervals.

$x < - 2$

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

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

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 $x < - 2$ to see if it makes the inequality true.

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

$x = - 4$

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

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

The left side $18$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

$2 {(0)}^{2} + 3 (0) - 2 > 0$

The left side $- 2$ is not greater than the right side $0$ , 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.

$2 {(3)}^{2} + 3 (3) - 2 > 0$

The left side $25$ 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.

$x < - 2$ True

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

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

$x < - 2$ True

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

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

The solution consists of all of the true intervals.

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

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

Solve the Inequality for x 2x^2+3x-2>0

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