Precalculus Examples

$x \leq 15 - 2 x^{2}$

Step 1

Add $2 x^{2}$ to both sides of the inequality.

$x + 2 x^{2} \leq 15$

Step 2

Convert the inequality to an equation.

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

Step 3

Subtract $15$ from both sides of the equation.

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

Step 4

Factor the left side of the equation.

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Step 4.1

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$u + 2 u^{2} - 15 = 0$

Step 4.2

Factor by grouping.

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Step 4.2.1

Reorder terms.

$2 u^{2} + u - 15 = 0$

Step 4.2.2

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 $b = 1$ .

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Step 4.2.2.1

Multiply by $1$ .

$2 u^{2} + 1 u - 15 = 0$

Step 4.2.2.2

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

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

Step 4.2.2.3

Apply the distributive property.

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

Step 4.2.3

Factor out the greatest common factor from each group.

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Step 4.2.3.1

Group the first two terms and the last two terms.

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

Step 4.2.3.2

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

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

Step 4.2.4

Factor the polynomial by factoring out the greatest common factor, $2 u - 5$ .

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

Step 4.3

Replace all occurrences of $u$ with $x$ .

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

Step 5

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

$2 x - 5 = 0$

$x + 3 = 0$

Step 6

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

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Step 6.1

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

$2 x - 5 = 0$

Step 6.2

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

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Step 6.2.1

Add $5$ to both sides of the equation.

$2 x = 5$

Step 6.2.2

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

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Step 6.2.2.1

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

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

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.

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

Step 6.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 7

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

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Step 7.1

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

$x + 3 = 0$

Step 7.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 8

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

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

Step 9

Use each root to create test intervals.

$x < - 3$

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

$x > \frac{5}{2}$

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

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Step 10.1.1

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

$x = - 6$

Step 10.1.2

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

$- 6 \leq 15 - 2 {(- 6)}^{2}$

Step 10.1.3

The left side $- 6$ is greater than the right side $- 57$ , which means that the given statement is false.

False

Step 10.2

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

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Step 10.2.1

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

$x = 0$

Step 10.2.2

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

$0 \leq 15 - 2 {(0)}^{2}$

Step 10.2.3

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

True

Step 10.3

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

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Step 10.3.1

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

$x = 5$

Step 10.3.2

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

$5 \leq 15 - 2 {(5)}^{2}$

Step 10.3.3

The left side $5$ is greater than the right side $- 35$ , which means that the given statement is false.

False

Step 10.4

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

$x < - 3$ False

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

$x > \frac{5}{2}$ False

$x < - 3$ False

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

$x > \frac{5}{2}$ False

Step 11

The solution consists of all of the true intervals.

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

Step 12

Convert the inequality to interval notation.

$[- 3, \frac{5}{2}]$

Step 13