Solve for x 2x^2+5x-3<0

$2 x^{2} + 5 x - 3 < 0$

Step 1

Convert the inequality to an equation.

$2 x^{2} + 5 x - 3 = 0$

Step 2

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 $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = 5$ .

Factor $5$ out of $5 x$ .

$2 x^{2} + 5 (x) - 3 = 0$

Rewrite $5$ as $- 1$ plus $6$

$2 x^{2} + (- 1 + 6) x - 3 = 0$

Apply the distributive property.

$2 x^{2} - 1 x + 6 x - 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{2} - 1 x) + 6 x - 3 = 0$

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

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

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

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

Step 3

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

$2 x - 1 = 0$

$x + 3 = 0$

Step 4

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 5

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 6

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

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

Step 7

Use each root to create test intervals.

$x < - 3$

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

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

Step 8

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 $x < - 3$ to see if it makes the inequality true.

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

$x = - 6$

Replace $x$ with $- 6$ in the original inequality.

$2 {(- 6)}^{2} + 5 (- 6) - 3 < 0$

The left side $39$ is not less than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 0$

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

$2 {(0)}^{2} + 5 (0) - 3 < 0$

The left side $- 3$ is less than the right side $0$ , which means that the given statement is always true.

True

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.

$2 {(3)}^{2} + 5 (3) - 3 < 0$

The left side $30$ is not less than the right side $0$ , which means that the given statement is false.

False

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

$x < - 3$ False

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

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

$x < - 3$ False

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

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

Step 9

The solution consists of all of the true intervals.

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

Step 10

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

Step 11

Solve for x 2x^2+5x-3<0

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