Precalculus Examples

$\sqrt{x^{2} - 2 x - 2} + \sqrt{2 x - 3} = 5$

Step 1

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

$x^{2} - 2 x - 2 \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Convert the inequality to an equation.

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

Step 2.2

Use the quadratic formula to find the solutions.

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

Step 2.3

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 2$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot - 2)}}{2 \cdot 1}$

Step 2.4

Simplify.

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Step 2.4.1

Simplify the numerator.

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Step 2.4.1.1

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 2}}{2 \cdot 1}$

Step 2.4.1.2

Multiply $- 4 \cdot 1 \cdot - 2$ .

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Step 2.4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 2}}{2 \cdot 1}$

Step 2.4.1.2.2

Multiply $- 4$ by $- 2$ .

$x = \frac{2 \pm \sqrt{4 + 8}}{2 \cdot 1}$

Step 2.4.1.3

Add $4$ and $8$ .

$x = \frac{2 \pm \sqrt{12}}{2 \cdot 1}$

Step 2.4.1.4

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 2.4.1.4.1

Factor $4$ out of $12$ .

$x = \frac{2 \pm \sqrt{4 (3)}}{2 \cdot 1}$

Step 2.4.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 2.4.1.5

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{3}}{2 \cdot 1}$

Step 2.4.2

Multiply $2$ by $1$ .

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

Step 2.4.3

Simplify $\frac{2 \pm 2 \sqrt{3}}{2}$ .

$x = 1 \pm \sqrt{3}$

Step 2.5

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

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Step 2.5.1

Simplify the numerator.

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Step 2.5.1.1

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 2}}{2 \cdot 1}$

Step 2.5.1.2

Multiply $- 4 \cdot 1 \cdot - 2$ .

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Step 2.5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 2}}{2 \cdot 1}$

Step 2.5.1.2.2

Multiply $- 4$ by $- 2$ .

$x = \frac{2 \pm \sqrt{4 + 8}}{2 \cdot 1}$

Step 2.5.1.3

Add $4$ and $8$ .

$x = \frac{2 \pm \sqrt{12}}{2 \cdot 1}$

Step 2.5.1.4

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 2.5.1.4.1

Factor $4$ out of $12$ .

$x = \frac{2 \pm \sqrt{4 (3)}}{2 \cdot 1}$

Step 2.5.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 2.5.1.5

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{3}}{2 \cdot 1}$

Step 2.5.2

Multiply $2$ by $1$ .

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

Step 2.5.3

Simplify $\frac{2 \pm 2 \sqrt{3}}{2}$ .

$x = 1 \pm \sqrt{3}$

Step 2.5.4

Change the $\pm$ to $+$ .

$x = 1 + \sqrt{3}$

Step 2.6

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

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Step 2.6.1

Simplify the numerator.

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Step 2.6.1.1

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 2}}{2 \cdot 1}$

Step 2.6.1.2

Multiply $- 4 \cdot 1 \cdot - 2$ .

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Step 2.6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 2}}{2 \cdot 1}$

Step 2.6.1.2.2

Multiply $- 4$ by $- 2$ .

$x = \frac{2 \pm \sqrt{4 + 8}}{2 \cdot 1}$

Step 2.6.1.3

Add $4$ and $8$ .

$x = \frac{2 \pm \sqrt{12}}{2 \cdot 1}$

Step 2.6.1.4

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 2.6.1.4.1

Factor $4$ out of $12$ .

$x = \frac{2 \pm \sqrt{4 (3)}}{2 \cdot 1}$

Step 2.6.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Step 2.6.1.5

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{3}}{2 \cdot 1}$

Step 2.6.2

Multiply $2$ by $1$ .

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

Step 2.6.3

Simplify $\frac{2 \pm 2 \sqrt{3}}{2}$ .

$x = 1 \pm \sqrt{3}$

Step 2.6.4

Change the $\pm$ to $-$ .

$x = 1 - \sqrt{3}$

Step 2.7

Consolidate the solutions.

$x = 1 + \sqrt{3}, 1 - \sqrt{3}$

Step 2.8

Use each root to create test intervals.

$x < 1 - \sqrt{3}$

$1 - \sqrt{3} < x < 1 + \sqrt{3}$

$x > 1 + \sqrt{3}$

Step 2.9

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 2.9.1

Test a value on the interval $x < 1 - \sqrt{3}$ to see if it makes the inequality true.

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Step 2.9.1.1

Choose a value on the interval $x < 1 - \sqrt{3}$ and see if this value makes the original inequality true.

$x = - 3$

Step 2.9.1.2

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

${(- 3)}^{2} - 2 \cdot - 3 - 2 \geq 0$

Step 2.9.1.3

The left side $13$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.9.2

Test a value on the interval $1 - \sqrt{3} < x < 1 + \sqrt{3}$ to see if it makes the inequality true.

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Step 2.9.2.1

Choose a value on the interval $1 - \sqrt{3} < x < 1 + \sqrt{3}$ and see if this value makes the original inequality true.

$x = 0$

Step 2.9.2.2

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

${(0)}^{2} - 2 \cdot 0 - 2 \geq 0$

Step 2.9.2.3

The left side $- 2$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.9.3

Test a value on the interval $x > 1 + \sqrt{3}$ to see if it makes the inequality true.

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Step 2.9.3.1

Choose a value on the interval $x > 1 + \sqrt{3}$ and see if this value makes the original inequality true.

$x = 5$

Step 2.9.3.2

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

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

Step 2.9.3.3

The left side $13$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.9.4

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

$x < 1 - \sqrt{3}$ True

$1 - \sqrt{3} < x < 1 + \sqrt{3}$ False

$x > 1 + \sqrt{3}$ True

$x < 1 - \sqrt{3}$ True

$1 - \sqrt{3} < x < 1 + \sqrt{3}$ False

$x > 1 + \sqrt{3}$ True

Step 2.10

The solution consists of all of the true intervals.

$x \leq 1 - \sqrt{3}$ or $x \geq 1 + \sqrt{3}$

Step 3

Set the radicand in $\sqrt{2 x - 3}$ greater than or equal to $0$ to find where the expression is defined.

$2 x - 3 \geq 0$

Step 4

Solve for $x$ .

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Step 4.1

Add $3$ to both sides of the inequality.

$2 x \geq 3$

Step 4.2

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

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Step 4.2.1

Divide each term in $2 x \geq 3$ by $2$ .

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

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $2$ .

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Step 4.2.2.1.1

Cancel the common factor.

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

Step 4.2.2.1.2

Divide $x$ by $1$ .

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

Step 5

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

Interval Notation:

$[1 + \sqrt{3}, \infty)$

Set-Builder Notation:

${x | x \geq 1 + \sqrt{3}}$

Step 6