Precalculus Examples

$2 x^{2} < 9 x + 5$

Step 1

Subtract $9 x$ from both sides of the inequality.

$2 x^{2} - 9 x < 5$

Step 2

Convert the inequality to an equation.

$2 x^{2} - 9 x = 5$

Step 3

Subtract $5$ from both sides of the equation.

$2 x^{2} - 9 x - 5 = 0$

Step 4

Factor by grouping.

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Step 4.1

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 $a \cdot c = 2 \cdot - 5 = - 10$ and whose sum is $b = - 9$ .

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Step 4.1.1

Factor $- 9$ out of $- 9 x$ .

$2 x^{2} - 9 x - 5 = 0$

Step 4.1.2

Rewrite $- 9$ as $1$ plus $- 10$

$2 x^{2} + (1 - 10) x - 5 = 0$

Step 4.1.3

Apply the distributive property.

$2 x^{2} + 1 x - 10 x - 5 = 0$

Step 4.1.4

Multiply $x$ by $1$ .

$2 x^{2} + x - 10 x - 5 = 0$

Step 4.2

Factor out the greatest common factor from each group.

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Step 4.2.1

Group the first two terms and the last two terms.

$(2 x^{2} + x) - 10 x - 5 = 0$

Step 4.2.2

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

$x (2 x + 1) - 5 (2 x + 1) = 0$

Step 4.3

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

$(2 x + 1) (x - 5) = 0$

Step 5

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

$2 x + 1 = 0$

$x - 5 = 0$

Step 6

Set $2 x + 1$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Step 6.2

Solve $2 x + 1 = 0$ for $x$ .

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Step 6.2.1

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Step 6.2.2

Divide each term in $2 x = - 1$ by $2$ and simplify.

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Step 6.2.2.1

Divide each term in $2 x = - 1$ by $2$ .

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

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of $2$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

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

Step 6.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 6.2.2.3

Simplify the right side.

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Step 6.2.2.3.1

Move the negative in front of the fraction.

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

Step 7

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

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Step 7.1

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

$x - 5 = 0$

Step 7.2

Add $5$ to both sides of the equation.

$x = 5$

Step 8

The final solution is all the values that make $(2 x + 1) (x - 5) = 0$ true.

$x = - \frac{1}{2}, 5$

Step 9

Use each root to create test intervals.

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

$- \frac{1}{2} < x < 5$

$x > 5$

Step 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 10.1

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

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Step 10.1.1

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

$x = - 3$

Step 10.1.2

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

$2 {(- 3)}^{2} < 9 (- 3) + 5$

Step 10.1.3

The left side $18$ is not less than the right side $- 22$ , which means that the given statement is false.

False

Step 10.2

Test a value on the interval $- \frac{1}{2} < x < 5$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $- \frac{1}{2} < x < 5$ and see if this value makes the original inequality true.

$x = 0$

Step 10.2.2

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

$2 {(0)}^{2} < 9 (0) + 5$

Step 10.2.3

The left side $0$ is less than the right side $5$ , which means that the given statement is always true.

True

Step 10.3

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

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Step 10.3.1

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

$x = 8$

Step 10.3.2

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

$2 {(8)}^{2} < 9 (8) + 5$

Step 10.3.3

The left side $128$ is not less than the right side $77$ , which means that the given statement is false.

False

Step 10.4

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

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

$- \frac{1}{2} < x < 5$ True

$x > 5$ False

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

$- \frac{1}{2} < x < 5$ True

$x > 5$ False

Step 11

The solution consists of all of the true intervals.

$- \frac{1}{2} < x < 5$

Step 12

Convert the inequality to interval notation.

$(- \frac{1}{2}, 5)$

Step 13