Solve the Inequality for x 2x^2<x+15

$2 x^{2} < x + 15$

Subtract $x$ from both sides of the inequality.

$2 x^{2} - x < 15$

Convert the inequality to an equation.

$2 x^{2} - x = 15$

Subtract $15$ from both sides of the equation.

$2 x^{2} - x - 15 = 0$

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 $a \cdot c = 2 \cdot - 15 = - 30$ and whose sum is $b = - 1$ .

Factor $- 1$ out of $- x$ .

$2 x^{2} - (x) - 15 = 0$

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

$2 x^{2} + (5 - 6) x - 15 = 0$

Apply the distributive property.

$2 x^{2} + 5 x - 6 x - 15 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{2} + 5 x) - 6 x - 15 = 0$

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

$x (2 x + 5) - 3 (2 x + 5) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x + 5$ .

$(2 x + 5) (x - 3) = 0$

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

$2 x + 5 = 0$

$x - 3 = 0$

Set $2 x + 5$ equal to $0$ and solve for $x$ .

Set $2 x + 5$ equal to $0$ .

$2 x + 5 = 0$

Solve $2 x + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Divide each term in $2 x = - 5$ by $2$ and simplify.

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

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 $(2 x + 5) (x - 3) = 0$ true.

$x = - \frac{5}{2}, 3$

Use each root to create test intervals.

$x < - \frac{5}{2}$

$- \frac{5}{2} < x < 3$

$x > 3$

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 $x < - \frac{5}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x < - \frac{5}{2}$ and see if this value makes the original inequality true.

$x = - 5$

Replace $x$ with $- 5$ in the original inequality.

$2 {(- 5)}^{2} < (- 5) + 15$

The left side $50$ is not less than the right side $10$ , which means that the given statement is false.

False

Test a value on the interval $- \frac{5}{2} < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{5}{2} < x < 3$ and see if this value makes the original inequality true.

$x = 0$

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

$2 {(0)}^{2} < (0) + 15$

The left side $0$ is less than the right side $15$ , 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.

$2 {(6)}^{2} < (6) + 15$

The left side $72$ is not less than the right side $21$ , which means that the given statement is false.

False

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

$x < - \frac{5}{2}$ False

$- \frac{5}{2} < x < 3$ True

$x > 3$ False

$x < - \frac{5}{2}$ False

$- \frac{5}{2} < x < 3$ True

$x > 3$ False

The solution consists of all of the true intervals.

$- \frac{5}{2} < x < 3$

The result can be shown in multiple forms.

Inequality Form:

$- \frac{5}{2} < x < 3$

Interval Notation:

$(- \frac{5}{2}, 3)$

Solve the Inequality for x 2x^2<x+15

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