Algebra Examples

$\sqrt{x^{2} - 6 x + 9}$

Step 1

Set the radicand in

$\sqrt{x^{2} - 6 x + 9}$ greater than or equal to

$0$ to find where the expression is defined.

$x^{2} - 6 x + 9 \geq 0$

Step 2

Solve for

$x$ .

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Step 2.1

Convert the inequality to an equation.

$x^{2} - 6 x + 9 = 0$

Step 2.2

Factor using the perfect square rule.

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Step 2.2.1

Rewrite

$9$ as

$3^{2}$ .

$x^{2} - 6 x + 3^{2} = 0$

Step 2.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Step 2.2.3

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 3 + 3^{2} = 0$

Step 2.2.4

Factor using the perfect square trinomial rule

$a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where

$a = x$ and

$b = 3$ .

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

Step 2.3

Set the

$x - 3$ equal to

$0$ .

$x - 3 = 0$

Step 2.4

Add

$3$ to both sides of the equation.

$x = 3$

Step 2.5

Use each root to create test intervals.

$x < 3$

$x > 3$

Step 2.6

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 2.6.1

Test a value on the interval

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

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Step 2.6.1.1

Choose a value on the interval

$x < 3$ and see if this value makes the original inequality true.

$x = 0$

Step 2.6.1.2

Replace

$x$ with

$0$ in the original inequality.

${(0)}^{2} - 6 \cdot 0 + 9 \geq 0$

Step 2.6.1.3

The left side

$9$ is greater than the right side

$0$ , which means that the given statement is always true.

True

Step 2.6.2

Test a value on the interval

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

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Step 2.6.2.1

Choose a value on the interval

$x > 3$ and see if this value makes the original inequality true.

$x = 6$

Step 2.6.2.2

Replace

$x$ with

$6$ in the original inequality.

${(6)}^{2} - 6 \cdot 6 + 9 \geq 0$

Step 2.6.2.3

The left side

$9$ is greater than the right side

$0$ , which means that the given statement is always true.

True

Step 2.6.3

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

$x < 3$ True

$x > 3$ True

$x < 3$ True

$x > 3$ True

Step 2.7

The solution consists of all of the true intervals.

$x \leq 3$ or

$x \geq 3$

Step 2.8

Combine the intervals.

All real numbers

Step 3

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 4