Solve for x 3+ square root of 3x+12<=9

$3 + \sqrt{3 x + 12} \leq 9$

Step 1

Move all terms not containing $\sqrt{3 x + 12}$ to the right side of the inequality.

Subtract $3$ from both sides of the inequality.

$\sqrt{3 x + 12} \leq 9 - 3$

Subtract $3$ from $9$ .

$\sqrt{3 x + 12} \leq 6$

Step 2

To remove the radical on the left side of the inequality, square both sides of the inequality.

${\sqrt{3 x + 12}}^{2} \leq 6^{2}$

Step 3

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 x + 12}$ as ${(3 x + 12)}^{\frac{1}{2}}$ .

${({(3 x + 12)}^{\frac{1}{2}})}^{2} \leq 6^{2}$

Simplify the left side.

Simplify ${({(3 x + 12)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(3 x + 12)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(3 x + 12)}^{\frac{1}{2} \cdot 2} \leq 6^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(3 x + 12)}^{\frac{1}{2} \cdot 2} \leq 6^{2}$

Rewrite the expression.

${(3 x + 12)}^{1} \leq 6^{2}$

Simplify.

$3 x + 12 \leq 6^{2}$

Simplify the right side.

Raise $6$ to the power of $2$ .

$3 x + 12 \leq 36$

Step 4

Solve for $x$ .

Move all terms not containing $x$ to the right side of the inequality.

Subtract $12$ from both sides of the inequality.

$3 x \leq 36 - 12$

Subtract $12$ from $36$ .

$3 x \leq 24$

Divide each term in $3 x \leq 24$ by $3$ and simplify.

Divide each term in $3 x \leq 24$ by $3$ .

$\frac{3 x}{3} \leq \frac{24}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \leq \frac{24}{3}$

Divide $x$ by $1$ .

$x \leq \frac{24}{3}$

Simplify the right side.

Divide $24$ by $3$ .

$x \leq 8$

Step 5

Find the domain of $- 6 + \sqrt{3 (x + 4)}$ .

Set the radicand in $\sqrt{3 (x + 4)}$ greater than or equal to $0$ to find where the expression is defined.

$3 (x + 4) \geq 0$

Solve for $x$ .

Divide each term in $3 (x + 4) \geq 0$ by $3$ and simplify.

Divide each term in $3 (x + 4) \geq 0$ by $3$ .

$\frac{3 (x + 4)}{3} \geq \frac{0}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 (x + 4)}{3} \geq \frac{0}{3}$

Divide $x + 4$ by $1$ .

$x + 4 \geq \frac{0}{3}$

Simplify the right side.

Divide $0$ by $3$ .

$x + 4 \geq 0$

Subtract $4$ from both sides of the inequality.

$x \geq - 4$

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

$[- 4, \infty)$

Step 6

Use each root to create test intervals.

$x < - 4$

$- 4 < x < 8$

$x > 8$

Step 7

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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

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

$x = - 6$

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

$3 + \sqrt{3 (- 6) + 12} \leq 9$

The left side is not equal to the right side, which means that the given statement is false.

False

Test a value on the interval $- 4 < x < 8$ to see if it makes the inequality true.

Choose a value on the interval $- 4 < x < 8$ and see if this value makes the original inequality true.

$x = 0$

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

$3 + \sqrt{3 (0) + 12} \leq 9$

The left side $6.46410161$ is less than the right side $9$ , which means that the given statement is always true.

True

Test a value on the interval $x > 8$ to see if it makes the inequality true.

Choose a value on the interval $x > 8$ and see if this value makes the original inequality true.

$x = 10$

Replace $x$ with $10$ in the original inequality.

$3 + \sqrt{3 (10) + 12} \leq 9$

The left side $9.48074069$ is greater than the right side $9$ , which means that the given statement is false.

False

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

$x < - 4$ False

$- 4 < x < 8$ True

$x > 8$ False

$x < - 4$ False

$- 4 < x < 8$ True

$x > 8$ False

Step 8

The solution consists of all of the true intervals.

$- 4 \leq x \leq 8$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$- 4 \leq x \leq 8$

Interval Notation:

$[- 4, 8]$

Step 10

Solve for x 3+ square root of 3x+12<=9

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