Convert to Set Notation 2x^2-5x-3<=0

$2 x^{2} - 5 x - 3 \leq 0$

Step 1

Solve $- \frac{1}{2} \leq x \leq 3$ .

Convert the inequality to an equation.

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

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 $a \cdot c = 2 \cdot - 3 = - 6$ 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.

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

Multiply $x$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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, $2 x + 1$ .

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

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

$2 x + 1 = 0$

$x - 3 = 0$

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.

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

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

Use each root to create test intervals.

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

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

$x > 3$

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

$2 {(- 3)}^{2} - 5 \cdot - 3 - 3 \leq 0$

The left side $30$ is greater than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 0$

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

$2 {(0)}^{2} - 5 \cdot 0 - 3 \leq 0$

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 $x > 3$ 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.

$2 {(6)}^{2} - 5 \cdot 6 - 3 \leq 0$

The left side $39$ is greater 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.

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

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

$x > 3$ False

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

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

$x > 3$ False

The solution consists of all of the true intervals.

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

Step 2

Use the inequality $- \frac{1}{2} \leq x \leq 3$ to build the set notation.

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

Step 3

Convert to Set Notation 2x^2-5x-3<=0

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