Precalculus Examples

$f (x) = \sqrt{2 x^{2} + 5 x - 12}$

Step 1

Set the radicand in $\sqrt{2 x^{2} + 5 x - 12}$ greater than or equal to $0$ to find where the expression is defined.

$2 x^{2} + 5 x - 12 \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Convert the inequality to an equation.

$2 x^{2} + 5 x - 12 = 0$

Step 2.2

Factor by grouping.

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Step 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 $a \cdot c = 2 \cdot - 12 = - 24$ and whose sum is $b = 5$ .

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Step 2.2.1.1

Factor $5$ out of $5 x$ .

$2 x^{2} + 5 (x) - 12 = 0$

Step 2.2.1.2

Rewrite $5$ as $- 3$ plus $8$

$2 x^{2} + (- 3 + 8) x - 12 = 0$

Step 2.2.1.3

Apply the distributive property.

$2 x^{2} - 3 x + 8 x - 12 = 0$

Step 2.2.2

Factor out the greatest common factor from each group.

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Step 2.2.2.1

Group the first two terms and the last two terms.

$(2 x^{2} - 3 x) + 8 x - 12 = 0$

Step 2.2.2.2

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

$x (2 x - 3) + 4 (2 x - 3) = 0$

Step 2.2.3

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

$(2 x - 3) (x + 4) = 0$

Step 2.3

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

$2 x - 3 = 0$

$x + 4 = 0$

Step 2.4

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

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Step 2.4.1

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

$2 x - 3 = 0$

Step 2.4.2

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

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Step 2.4.2.1

Add $3$ to both sides of the equation.

$2 x = 3$

Step 2.4.2.2

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

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Step 2.4.2.2.1

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

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

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

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

Step 2.4.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 2.5

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 2.5.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 2.6

The final solution is all the values that make $(2 x - 3) (x + 4) = 0$ true.

$x = \frac{3}{2}, - 4$

Step 2.7

Use each root to create test intervals.

$x < - 4$

$- 4 < x < \frac{3}{2}$

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

Step 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 2.8.1

Test a value on the interval $x < - 4$ to see if it makes the inequality true.

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Step 2.8.1.1

Choose a value on the interval $x < - 4$ and see if this value makes the original inequality true.

$x = - 6$

Step 2.8.1.2

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

$2 {(- 6)}^{2} + 5 (- 6) - 12 \geq 0$

Step 2.8.1.3

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

True

Step 2.8.2

Test a value on the interval $- 4 < x < \frac{3}{2}$ to see if it makes the inequality true.

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Step 2.8.2.1

Choose a value on the interval $- 4 < x < \frac{3}{2}$ and see if this value makes the original inequality true.

$x = 0$

Step 2.8.2.2

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

$2 {(0)}^{2} + 5 (0) - 12 \geq 0$

Step 2.8.2.3

The left side $- 12$ is less than the right side $0$ , which means that the given statement is false.

False

Step 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 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 2.8.3.2

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

$2 {(4)}^{2} + 5 (4) - 12 \geq 0$

Step 2.8.3.3

The left side $40$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.8.4

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

$x < - 4$ True

$- 4 < x < \frac{3}{2}$ False

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

$x < - 4$ True

$- 4 < x < \frac{3}{2}$ False

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

Step 2.9

The solution consists of all of the true intervals.

$x \leq - 4$ or $x \geq \frac{3}{2}$

Step 3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, - 4] \cup [\frac{3}{2}, \infty)$

Set-Builder Notation:

${x | x \leq - 4, x \geq \frac{3}{2}}$

Step 4