Precalculus Examples

$| 2 x^{2} + 7 x - 15 | < 10$

Step 1

Write $| 2 x^{2} + 7 x - 15 | < 10$ as a piecewise.

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Step 1.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$2 x^{2} + 7 x - 15 \geq 0$

Step 1.2

Solve the inequality.

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Step 1.2.1

Convert the inequality to an equation.

$2 x^{2} + 7 x - 15 = 0$

Step 1.2.2

Factor by grouping.

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Step 1.2.2.1

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 $b = 7$ .

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Step 1.2.2.1.1

Factor $7$ out of $7 x$ .

$2 x^{2} + 7 (x) - 15 = 0$

Step 1.2.2.1.2

Rewrite $7$ as $- 3$ plus $10$

$2 x^{2} + (- 3 + 10) x - 15 = 0$

Step 1.2.2.1.3

Apply the distributive property.

$2 x^{2} - 3 x + 10 x - 15 = 0$

Step 1.2.2.2

Factor out the greatest common factor from each group.

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Step 1.2.2.2.1

Group the first two terms and the last two terms.

$(2 x^{2} - 3 x) + 10 x - 15 = 0$

Step 1.2.2.2.2

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

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

Step 1.2.2.3

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

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

Step 1.2.3

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

$2 x - 3 = 0$

$x + 5 = 0$

Step 1.2.4

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

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Step 1.2.4.1

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

$2 x - 3 = 0$

Step 1.2.4.2

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

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Step 1.2.4.2.1

Add $3$ to both sides of the equation.

$2 x = 3$

Step 1.2.4.2.2

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

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Step 1.2.4.2.2.1

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

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

Step 1.2.4.2.2.2

Simplify the left side.

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Step 1.2.4.2.2.2.1

Cancel the common factor of $2$ .

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Step 1.2.4.2.2.2.1.1

Cancel the common factor.

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

Step 1.2.4.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 1.2.5

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

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Step 1.2.5.1

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

$x + 5 = 0$

Step 1.2.5.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 1.2.6

The final solution is all the values that make $(2 x - 3) (x + 5) = 0$ true.

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

Step 1.2.7

Use each root to create test intervals.

$x < - 5$

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

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

Step 1.2.8

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 1.2.8.1

Test a value on the interval $x < - 5$ to see if it makes the inequality true.

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Step 1.2.8.1.1

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

$x = - 8$

Step 1.2.8.1.2

Replace $x$ with $- 8$ in the original inequality.

$2 {(- 8)}^{2} + 7 (- 8) - 15 \geq 0$

Step 1.2.8.1.3

The left side $57$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.2.8.2

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

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Step 1.2.8.2.1

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

$x = 0$

Step 1.2.8.2.2

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

$2 {(0)}^{2} + 7 (0) - 15 \geq 0$

Step 1.2.8.2.3

The left side $- 15$ is less than the right side $0$ , which means that the given statement is false.

False

Step 1.2.8.3

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

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Step 1.2.8.3.1

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

$x = 4$

Step 1.2.8.3.2

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

$2 {(4)}^{2} + 7 (4) - 15 \geq 0$

Step 1.2.8.3.3

The left side $45$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.2.8.4

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

$x < - 5$ True

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

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

$x < - 5$ True

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

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

Step 1.2.9

The solution consists of all of the true intervals.

$x \leq - 5$ or $x \geq \frac{3}{2}$

Step 1.3

In the piece where $2 x^{2} + 7 x - 15$ is non-negative, remove the absolute value.

$2 x^{2} + 7 x - 15 < 10$

Step 1.4

To find the interval for the second piece, find where the inside of the absolute value is negative.

$2 x^{2} + 7 x - 15 < 0$

Step 1.5

Solve the inequality.

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Step 1.5.1

Convert the inequality to an equation.

$2 x^{2} + 7 x - 15 = 0$

Step 1.5.2

Factor by grouping.

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Step 1.5.2.1

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 $b = 7$ .

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Step 1.5.2.1.1

Factor $7$ out of $7 x$ .

$2 x^{2} + 7 (x) - 15 = 0$

Step 1.5.2.1.2

Rewrite $7$ as $- 3$ plus $10$

$2 x^{2} + (- 3 + 10) x - 15 = 0$

Step 1.5.2.1.3

Apply the distributive property.

$2 x^{2} - 3 x + 10 x - 15 = 0$

Step 1.5.2.2

Factor out the greatest common factor from each group.

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Step 1.5.2.2.1

Group the first two terms and the last two terms.

$(2 x^{2} - 3 x) + 10 x - 15 = 0$

Step 1.5.2.2.2

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

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

Step 1.5.2.3

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

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

Step 1.5.3

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

$2 x - 3 = 0$

$x + 5 = 0$

Step 1.5.4

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

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Step 1.5.4.1

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

$2 x - 3 = 0$

Step 1.5.4.2

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

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Step 1.5.4.2.1

Add $3$ to both sides of the equation.

$2 x = 3$

Step 1.5.4.2.2

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

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Step 1.5.4.2.2.1

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

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

Step 1.5.4.2.2.2

Simplify the left side.

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Step 1.5.4.2.2.2.1

Cancel the common factor of $2$ .

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Step 1.5.4.2.2.2.1.1

Cancel the common factor.

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

Step 1.5.4.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 1.5.5

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

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Step 1.5.5.1

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

$x + 5 = 0$

Step 1.5.5.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 1.5.6

The final solution is all the values that make $(2 x - 3) (x + 5) = 0$ true.

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

Step 1.5.7

Use each root to create test intervals.

$x < - 5$

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

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

Step 1.5.8

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 1.5.8.1

Test a value on the interval $x < - 5$ to see if it makes the inequality true.

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Step 1.5.8.1.1

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

$x = - 8$

Step 1.5.8.1.2

Replace $x$ with $- 8$ in the original inequality.

$2 {(- 8)}^{2} + 7 (- 8) - 15 < 0$

Step 1.5.8.1.3

The left side $57$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 1.5.8.2

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

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Step 1.5.8.2.1

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

$x = 0$

Step 1.5.8.2.2

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

$2 {(0)}^{2} + 7 (0) - 15 < 0$

Step 1.5.8.2.3

The left side $- 15$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 1.5.8.3

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

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Step 1.5.8.3.1

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

$x = 4$

Step 1.5.8.3.2

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

$2 {(4)}^{2} + 7 (4) - 15 < 0$

Step 1.5.8.3.3

The left side $45$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 1.5.8.4

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

$x < - 5$ False

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

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

$x < - 5$ False

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

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

Step 1.5.9

The solution consists of all of the true intervals.

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

Step 1.6

In the piece where $2 x^{2} + 7 x - 15$ is negative, remove the absolute value and multiply by $- 1$ .

$- (2 x^{2} + 7 x - 15) < 10$

Step 1.7

Write as a piecewise.

${\begin{cases} 2 x^{2} + 7 x - 15 < 10 & x \leq - 5 or x \geq \frac{3}{2} \\ - (2 x^{2} + 7 x - 15) < 10 & - 5 < x < \frac{3}{2} \end{cases}$

Step 1.8

Simplify $- (2 x^{2} + 7 x - 15) < 10$ .

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Step 1.8.1

Apply the distributive property.

${\begin{cases} 2 x^{2} + 7 x - 15 < 10 & x \leq - 5 or x \geq \frac{3}{2} \\ - (2 x^{2}) - (7 x) - - 15 < 10 & - 5 < x < \frac{3}{2} \end{cases}$

Step 1.8.2

Simplify.

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Step 1.8.2.1

Multiply $2$ by $- 1$ .

${\begin{cases} 2 x^{2} + 7 x - 15 < 10 & x \leq - 5 or x \geq \frac{3}{2} \\ - 2 x^{2} - (7 x) - - 15 < 10 & - 5 < x < \frac{3}{2} \end{cases}$

Step 1.8.2.2

Multiply $7$ by $- 1$ .

${\begin{cases} 2 x^{2} + 7 x - 15 < 10 & x \leq - 5 or x \geq \frac{3}{2} \\ - 2 x^{2} - 7 x - - 15 < 10 & - 5 < x < \frac{3}{2} \end{cases}$

Step 1.8.2.3

Multiply $- 1$ by $- 15$ .

${\begin{cases} 2 x^{2} + 7 x - 15 < 10 & x \leq - 5 or x \geq \frac{3}{2} \\ - 2 x^{2} - 7 x + 15 < 10 & - 5 < x < \frac{3}{2} \end{cases}$

Step 2

Solve $2 x^{2} + 7 x - 15 < 10$ when $x \leq - 5 or x \geq \frac{3}{2}$ .

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Step 2.1

Solve $2 x^{2} + 7 x - 15 < 10$ for $x$ .

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Step 2.1.1

Move all terms to the left side of the equation and simplify.

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Step 2.1.1.1

Subtract $10$ from both sides of the inequality.

$2 x^{2} + 7 x - 15 - 10 < 0$

Step 2.1.1.2

Subtract $10$ from $- 15$ .

$2 x^{2} + 7 x - 25 < 0$

Step 2.1.2

Convert the inequality to an equation.

$2 x^{2} + 7 x - 25 = 0$

Step 2.1.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2.1.4

Substitute the values $a = 2$ , $b = 7$ , and $c = - 25$ into the quadratic formula and solve for $x$ .

$\frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (2 \cdot - 25)}}{2 \cdot 2}$

Step 2.1.5

Simplify.

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Step 2.1.5.1

Simplify the numerator.

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Step 2.1.5.1.1

Raise $7$ to the power of $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot 2 \cdot - 25}}{2 \cdot 2}$

Step 2.1.5.1.2

Multiply $- 4 \cdot 2 \cdot - 25$ .

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Step 2.1.5.1.2.1

Multiply $- 4$ by $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 8 \cdot - 25}}{2 \cdot 2}$

Step 2.1.5.1.2.2

Multiply $- 8$ by $- 25$ .

$x = \frac{- 7 \pm \sqrt{49 + 200}}{2 \cdot 2}$

Step 2.1.5.1.3

Add $49$ and $200$ .

$x = \frac{- 7 \pm \sqrt{249}}{2 \cdot 2}$

Step 2.1.5.2

Multiply $2$ by $2$ .

$x = \frac{- 7 \pm \sqrt{249}}{4}$

Step 2.1.6

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 2.1.6.1

Simplify the numerator.

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Step 2.1.6.1.1

Raise $7$ to the power of $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot 2 \cdot - 25}}{2 \cdot 2}$

Step 2.1.6.1.2

Multiply $- 4 \cdot 2 \cdot - 25$ .

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Step 2.1.6.1.2.1

Multiply $- 4$ by $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 8 \cdot - 25}}{2 \cdot 2}$

Step 2.1.6.1.2.2

Multiply $- 8$ by $- 25$ .

$x = \frac{- 7 \pm \sqrt{49 + 200}}{2 \cdot 2}$

Step 2.1.6.1.3

Add $49$ and $200$ .

$x = \frac{- 7 \pm \sqrt{249}}{2 \cdot 2}$

Step 2.1.6.2

Multiply $2$ by $2$ .

$x = \frac{- 7 \pm \sqrt{249}}{4}$

Step 2.1.6.3

Change the $\pm$ to $+$ .

$x = \frac{- 7 + \sqrt{249}}{4}$

Step 2.1.6.4

Rewrite $- 7$ as $- 1 (7)$ .

$x = \frac{- 1 \cdot 7 + \sqrt{249}}{4}$

Step 2.1.6.5

Factor $- 1$ out of $\sqrt{249}$ .

$x = \frac{- 1 \cdot 7 - 1 (- \sqrt{249})}{4}$

Step 2.1.6.6

Factor $- 1$ out of $- 1 (7) - 1 (- \sqrt{249})$ .

$x = \frac{- 1 (7 - \sqrt{249})}{4}$

Step 2.1.6.7

Move the negative in front of the fraction.

$x = - \frac{7 - \sqrt{249}}{4}$

Step 2.1.7

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 2.1.7.1

Simplify the numerator.

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Step 2.1.7.1.1

Raise $7$ to the power of $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot 2 \cdot - 25}}{2 \cdot 2}$

Step 2.1.7.1.2

Multiply $- 4 \cdot 2 \cdot - 25$ .

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Step 2.1.7.1.2.1

Multiply $- 4$ by $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 8 \cdot - 25}}{2 \cdot 2}$

Step 2.1.7.1.2.2

Multiply $- 8$ by $- 25$ .

$x = \frac{- 7 \pm \sqrt{49 + 200}}{2 \cdot 2}$

Step 2.1.7.1.3

Add $49$ and $200$ .

$x = \frac{- 7 \pm \sqrt{249}}{2 \cdot 2}$

Step 2.1.7.2

Multiply $2$ by $2$ .

$x = \frac{- 7 \pm \sqrt{249}}{4}$

Step 2.1.7.3

Change the $\pm$ to $-$ .

$x = \frac{- 7 - \sqrt{249}}{4}$

Step 2.1.7.4

Rewrite $- 7$ as $- 1 (7)$ .

$x = \frac{- 1 \cdot 7 - \sqrt{249}}{4}$

Step 2.1.7.5

Factor $- 1$ out of $- \sqrt{249}$ .

$x = \frac{- 1 \cdot 7 - (\sqrt{249})}{4}$

Step 2.1.7.6

Factor $- 1$ out of $- 1 (7) - (\sqrt{249})$ .

$x = \frac{- 1 (7 + \sqrt{249})}{4}$

Step 2.1.7.7

Move the negative in front of the fraction.

$x = - \frac{7 + \sqrt{249}}{4}$

Step 2.1.8

Consolidate the solutions.

$x = - \frac{7 - \sqrt{249}}{4}, - \frac{7 + \sqrt{249}}{4}$

Step 2.1.9

Use each root to create test intervals.

$x < - \frac{7 + \sqrt{249}}{4}$

$- \frac{7 + \sqrt{249}}{4} < x < - \frac{7 - \sqrt{249}}{4}$

$x > - \frac{7 - \sqrt{249}}{4}$

Step 2.1.10

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 2.1.10.1

Test a value on the interval $x < - \frac{7 + \sqrt{249}}{4}$ to see if it makes the inequality true.

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Step 2.1.10.1.1

Choose a value on the interval $x < - \frac{7 + \sqrt{249}}{4}$ and see if this value makes the original inequality true.

$x = - 8$

Step 2.1.10.1.2

Replace $x$ with $- 8$ in the original inequality.

$2 {(- 8)}^{2} + 7 (- 8) - 15 < 10$

Step 2.1.10.1.3

The left side $57$ is not less than the right side $10$ , which means that the given statement is false.

False

Step 2.1.10.2

Test a value on the interval $- \frac{7 + \sqrt{249}}{4} < x < - \frac{7 - \sqrt{249}}{4}$ to see if it makes the inequality true.

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Step 2.1.10.2.1

Choose a value on the interval $- \frac{7 + \sqrt{249}}{4} < x < - \frac{7 - \sqrt{249}}{4}$ and see if this value makes the original inequality true.

$x = 0$

Step 2.1.10.2.2

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

$2 {(0)}^{2} + 7 (0) - 15 < 10$

Step 2.1.10.2.3

The left side $- 15$ is less than the right side $10$ , which means that the given statement is always true.

True

Step 2.1.10.3

Test a value on the interval $x > - \frac{7 - \sqrt{249}}{4}$ to see if it makes the inequality true.

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Step 2.1.10.3.1

Choose a value on the interval $x > - \frac{7 - \sqrt{249}}{4}$ and see if this value makes the original inequality true.

$x = 5$

Step 2.1.10.3.2

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

$2 {(5)}^{2} + 7 (5) - 15 < 10$

Step 2.1.10.3.3

The left side $70$ is not less than the right side $10$ , which means that the given statement is false.

False

Step 2.1.10.4

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

$x < - \frac{7 + \sqrt{249}}{4}$ False

$- \frac{7 + \sqrt{249}}{4} < x < - \frac{7 - \sqrt{249}}{4}$ True

$x > - \frac{7 - \sqrt{249}}{4}$ False

$x < - \frac{7 + \sqrt{249}}{4}$ False

$- \frac{7 + \sqrt{249}}{4} < x < - \frac{7 - \sqrt{249}}{4}$ True

$x > - \frac{7 - \sqrt{249}}{4}$ False

Step 2.1.11

The solution consists of all of the true intervals.

$- \frac{7 + \sqrt{249}}{4} < x < - \frac{7 - \sqrt{249}}{4}$

Step 2.2

Find the intersection of $- \frac{7 + \sqrt{249}}{4} < x < - \frac{7 - \sqrt{249}}{4}$ and $x \leq - 5 o r + x \geq \frac{3}{2}$ .

$- \frac{7 + \sqrt{249}}{4} < x \leq - 5$ or $\frac{3}{2} \leq x < - \frac{7 - \sqrt{249}}{4}$

Step 3

Solve $- 2 x^{2} - 7 x + 15 < 10$ when $- 5 < x < \frac{3}{2}$ .

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Step 3.1

Solve $- 2 x^{2} - 7 x + 15 < 10$ for $x$ .

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Step 3.1.1

Move all terms to the left side of the equation and simplify.

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Step 3.1.1.1

Subtract $10$ from both sides of the inequality.

$- 2 x^{2} - 7 x + 15 - 10 < 0$

Step 3.1.1.2

Subtract $10$ from $15$ .

$- 2 x^{2} - 7 x + 5 < 0$

Step 3.1.2

Convert the inequality to an equation.

$- 2 x^{2} - 7 x + 5 = 0$

Step 3.1.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3.1.4

Substitute the values $a = - 2$ , $b = - 7$ , and $c = 5$ into the quadratic formula and solve for $x$ .

$\frac{7 \pm \sqrt{{(- 7)}^{2} - 4 \cdot (- 2 \cdot 5)}}{2 \cdot - 2}$

Step 3.1.5

Simplify.

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Step 3.1.5.1

Simplify the numerator.

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Step 3.1.5.1.1

Raise $- 7$ to the power of $2$ .

$x = \frac{7 \pm \sqrt{49 - 4 \cdot - 2 \cdot 5}}{2 \cdot - 2}$

Step 3.1.5.1.2

Multiply $- 4 \cdot - 2 \cdot 5$ .

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Step 3.1.5.1.2.1

Multiply $- 4$ by $- 2$ .

$x = \frac{7 \pm \sqrt{49 + 8 \cdot 5}}{2 \cdot - 2}$

Step 3.1.5.1.2.2

Multiply $8$ by $5$ .

$x = \frac{7 \pm \sqrt{49 + 40}}{2 \cdot - 2}$

Step 3.1.5.1.3

Add $49$ and $40$ .

$x = \frac{7 \pm \sqrt{89}}{2 \cdot - 2}$

Step 3.1.5.2

Multiply $2$ by $- 2$ .

$x = \frac{7 \pm \sqrt{89}}{- 4}$

Step 3.1.5.3

Move the negative in front of the fraction.

$x = - \frac{7 \pm \sqrt{89}}{4}$

Step 3.1.6

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 3.1.6.1

Simplify the numerator.

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Step 3.1.6.1.1

Raise $- 7$ to the power of $2$ .

$x = \frac{7 \pm \sqrt{49 - 4 \cdot - 2 \cdot 5}}{2 \cdot - 2}$

Step 3.1.6.1.2

Multiply $- 4 \cdot - 2 \cdot 5$ .

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Step 3.1.6.1.2.1

Multiply $- 4$ by $- 2$ .

$x = \frac{7 \pm \sqrt{49 + 8 \cdot 5}}{2 \cdot - 2}$

Step 3.1.6.1.2.2

Multiply $8$ by $5$ .

$x = \frac{7 \pm \sqrt{49 + 40}}{2 \cdot - 2}$

Step 3.1.6.1.3

Add $49$ and $40$ .

$x = \frac{7 \pm \sqrt{89}}{2 \cdot - 2}$

Step 3.1.6.2

Multiply $2$ by $- 2$ .

$x = \frac{7 \pm \sqrt{89}}{- 4}$

Step 3.1.6.3

Move the negative in front of the fraction.

$x = - \frac{7 \pm \sqrt{89}}{4}$

Step 3.1.6.4

Change the $\pm$ to $+$ .

$x = - \frac{7 + \sqrt{89}}{4}$

Step 3.1.7

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 3.1.7.1

Simplify the numerator.

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Step 3.1.7.1.1

Raise $- 7$ to the power of $2$ .

$x = \frac{7 \pm \sqrt{49 - 4 \cdot - 2 \cdot 5}}{2 \cdot - 2}$

Step 3.1.7.1.2

Multiply $- 4 \cdot - 2 \cdot 5$ .

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Step 3.1.7.1.2.1

Multiply $- 4$ by $- 2$ .

$x = \frac{7 \pm \sqrt{49 + 8 \cdot 5}}{2 \cdot - 2}$

Step 3.1.7.1.2.2

Multiply $8$ by $5$ .

$x = \frac{7 \pm \sqrt{49 + 40}}{2 \cdot - 2}$

Step 3.1.7.1.3

Add $49$ and $40$ .

$x = \frac{7 \pm \sqrt{89}}{2 \cdot - 2}$

Step 3.1.7.2

Multiply $2$ by $- 2$ .

$x = \frac{7 \pm \sqrt{89}}{- 4}$

Step 3.1.7.3

Move the negative in front of the fraction.

$x = - \frac{7 \pm \sqrt{89}}{4}$

Step 3.1.7.4

Change the $\pm$ to $-$ .

$x = - \frac{7 - \sqrt{89}}{4}$

Step 3.1.8

Consolidate the solutions.

$x = - \frac{7 + \sqrt{89}}{4}, - \frac{7 - \sqrt{89}}{4}$

Step 3.1.9

Use each root to create test intervals.

$x < - \frac{7 + \sqrt{89}}{4}$

$- \frac{7 + \sqrt{89}}{4} < x < - \frac{7 - \sqrt{89}}{4}$

$x > - \frac{7 - \sqrt{89}}{4}$

Step 3.1.10

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 3.1.10.1

Test a value on the interval $x < - \frac{7 + \sqrt{89}}{4}$ to see if it makes the inequality true.

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Step 3.1.10.1.1

Choose a value on the interval $x < - \frac{7 + \sqrt{89}}{4}$ and see if this value makes the original inequality true.

$x = - 7$

Step 3.1.10.1.2

Replace $x$ with $- 7$ in the original inequality.

$- 2 {(- 7)}^{2} - 7 \cdot - 7 + 15 < 10$

Step 3.1.10.1.3

The left side $- 34$ is less than the right side $10$ , which means that the given statement is always true.

True

Step 3.1.10.2

Test a value on the interval $- \frac{7 + \sqrt{89}}{4} < x < - \frac{7 - \sqrt{89}}{4}$ to see if it makes the inequality true.

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Step 3.1.10.2.1

Choose a value on the interval $- \frac{7 + \sqrt{89}}{4} < x < - \frac{7 - \sqrt{89}}{4}$ and see if this value makes the original inequality true.

$x = 0$

Step 3.1.10.2.2

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

$- 2 {(0)}^{2} - 7 \cdot 0 + 15 < 10$

Step 3.1.10.2.3

The left side $15$ is not less than the right side $10$ , which means that the given statement is false.

False

Step 3.1.10.3

Test a value on the interval $x > - \frac{7 - \sqrt{89}}{4}$ to see if it makes the inequality true.

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Step 3.1.10.3.1

Choose a value on the interval $x > - \frac{7 - \sqrt{89}}{4}$ and see if this value makes the original inequality true.

$x = 3$

Step 3.1.10.3.2

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

$- 2 {(3)}^{2} - 7 \cdot 3 + 15 < 10$

Step 3.1.10.3.3

The left side $- 24$ is less than the right side $10$ , which means that the given statement is always true.

True

Step 3.1.10.4

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

$x < - \frac{7 + \sqrt{89}}{4}$ True

$- \frac{7 + \sqrt{89}}{4} < x < - \frac{7 - \sqrt{89}}{4}$ False

$x > - \frac{7 - \sqrt{89}}{4}$ True

$x < - \frac{7 + \sqrt{89}}{4}$ True

$- \frac{7 + \sqrt{89}}{4} < x < - \frac{7 - \sqrt{89}}{4}$ False

$x > - \frac{7 - \sqrt{89}}{4}$ True

Step 3.1.11

The solution consists of all of the true intervals.

$x < - \frac{7 + \sqrt{89}}{4}$ or $x > - \frac{7 - \sqrt{89}}{4}$

Step 3.2

Find the intersection of $x < - \frac{7 + \sqrt{89}}{4} or x > - \frac{7 - \sqrt{89}}{4}$ and $- 5 < x < \frac{3}{2}$ .

$- 5 < x < - \frac{7 + \sqrt{89}}{4}$ or $- \frac{7 - \sqrt{89}}{4} < x < \frac{3}{2}$

Step 4

Find the union of the solutions.

$- \frac{7 + \sqrt{249}}{4} < x < - \frac{7 + \sqrt{89}}{4}$ or $- \frac{7 - \sqrt{89}}{4} < x < - \frac{7 - \sqrt{249}}{4}$

Step 5

Convert the inequality to interval notation.

$(- \frac{7 + \sqrt{249}}{4}, - \frac{7 + \sqrt{89}}{4}) \cup (- \frac{7 - \sqrt{89}}{4}, - \frac{7 - \sqrt{249}}{4})$

Step 6