Precalculus Examples

$\sqrt{x^{2} - 5 x + 6}$

Step 1

Set the radicand in $\sqrt{x^{2} - 5 x + 6}$ greater than or equal to $0$ to find where the expression is defined.

$x^{2} - 5 x + 6 \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Convert the inequality to an equation.

$x^{2} - 5 x + 6 = 0$

Step 2.2

Factor $x^{2} - 5 x + 6$ using the AC method.

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Step 2.2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $- 5$ .

$- 3, - 2$

Step 2.2.2

Write the factored form using these integers.

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

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 $0$ .

$x - 3 = 0$

$x - 2 = 0$

Step 2.4

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

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Step 2.4.1

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

$x - 3 = 0$

Step 2.4.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.5

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

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Step 2.5.1

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

$x - 2 = 0$

Step 2.5.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.6

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

$x = 3, 2$

Step 2.7

Use each root to create test intervals.

$x < 2$

$2 < x < 3$

$x > 3$

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 $x < 2$ to see if it makes the inequality true.

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Step 2.8.1.1

Choose a value on the interval $x < 2$ and see if this value makes the original inequality true.

$x = 0$

Step 2.8.1.2

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

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

Step 2.8.1.3

The left side $6$ 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 $2 < x < 3$ to see if it makes the inequality true.

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Step 2.8.2.1

Choose a value on the interval $2 < x < 3$ and see if this value makes the original inequality true.

$x = 2.5$

Step 2.8.2.2

Replace $x$ with $2.5$ in the original inequality.

${(2.5)}^{2} - 5 \cdot 2.5 + 6 \geq 0$

Step 2.8.2.3

The left side $- 0.25$ 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 $x > 3$ to see if it makes the inequality true.

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Step 2.8.3.1

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

$x = 6$

Step 2.8.3.2

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

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

Step 2.8.3.3

The left side $12$ 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.

$x < 2$ True

$2 < x < 3$ False

$x > 3$ True

$x < 2$ True

$2 < x < 3$ False

$x > 3$ True

Step 2.9

The solution consists of all of the true intervals.

$x \leq 2$ or $x \geq 3$

Step 3

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

Interval Notation:

$(- \infty, 2] \cup [3, \infty)$

Set-Builder Notation:

${x | x \leq 2, x \geq 3}$

Step 4