Precalculus Examples

$\sqrt{4 x^{2} - 12 x + 9}$

Step 1

Set the radicand in $\sqrt{4 x^{2} - 12 x + 9}$ greater than or equal to $0$ to find where the expression is defined.

$4 x^{2} - 12 x + 9 \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Convert the inequality to an equation.

$4 x^{2} - 12 x + 9 = 0$

Step 2.2

Factor using the perfect square rule.

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Step 2.2.1

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

${(2 x)}^{2} - 12 x + 9 = 0$

Step 2.2.2

Rewrite $9$ as $3^{2}$ .

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

Step 2.2.3

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

$12 x = 2 \cdot (2 x) \cdot 3$

Step 2.2.4

Rewrite the polynomial.

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

Step 2.2.5

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 2 x$ and $b = 3$ .

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

Step 2.3

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

$2 x - 3 = 0$

Step 2.4

Solve for $x$ .

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Step 2.4.1

Add $3$ to both sides of the equation.

$2 x = 3$

Step 2.4.2

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

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Step 2.4.2.1

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

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

Step 2.4.2.2

Simplify the left side.

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Step 2.4.2.2.1

Cancel the common factor of $2$ .

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Step 2.4.2.2.1.1

Cancel the common factor.

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

Step 2.4.2.2.1.2

Divide $x$ by $1$ .

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

Step 2.5

Use each root to create test intervals.

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

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

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 $x < \frac{3}{2}$ to see if it makes the inequality true.

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Step 2.6.1.1

Choose a value on the interval $x < \frac{3}{2}$ 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.

$4 {(0)}^{2} - 12 \cdot 0 + 9 \geq 0$

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 $x > \frac{3}{2}$ to see if it makes the inequality true.

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Step 2.6.2.1

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

$x = 4$

Step 2.6.2.2

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

$4 {(4)}^{2} - 12 \cdot 4 + 9 \geq 0$

Step 2.6.2.3

The left side $25$ 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.

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

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

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

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

Step 2.7

The solution consists of all of the true intervals.

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

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