Solve for x 2 square root of 3x+6-8<=4

$2 \sqrt{3 x + 6} - 8 \leq 4$

Move all terms not containing $2 \sqrt{3 x + 6}$ to the right side of the inequality.

Add $8$ to both sides of the inequality.

$2 \sqrt{3 x + 6} \leq 4 + 8$

Add $4$ and $8$ .

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

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

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

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 x + 6}$ as ${(3 x + 6)}^{\frac{1}{2}}$ .

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

Simplify the left side.

Simplify ${(2 {(3 x + 6)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $2 {(3 x + 6)}^{\frac{1}{2}}$ .

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

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

$4 {({(3 x + 6)}^{\frac{1}{2}})}^{2} \leq 12^{2}$

Multiply the exponents in ${({(3 x + 6)}^{\frac{1}{2}})}^{2}$ .

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

$4 {(3 x + 6)}^{\frac{1}{2} \cdot 2} \leq 12^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 {(3 x + 6)}^{\frac{1}{2} \cdot 2} \leq 12^{2}$

Rewrite the expression.

$4 {(3 x + 6)}^{1} \leq 12^{2}$

Simplify.

$4 (3 x + 6) \leq 12^{2}$

Apply the distributive property.

$4 (3 x) + 4 \cdot 6 \leq 12^{2}$

Multiply.

Multiply $3$ by $4$ .

$12 x + 4 \cdot 6 \leq 12^{2}$

Multiply $4$ by $6$ .

$12 x + 24 \leq 12^{2}$

Simplify the right side.

Raise $12$ to the power of $2$ .

$12 x + 24 \leq 144$

Solve for $x$ .

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

Subtract $24$ from both sides of the inequality.

$12 x \leq 144 - 24$

Subtract $24$ from $144$ .

$12 x \leq 120$

Divide each term in $12 x \leq 120$ by $12$ and simplify.

Divide each term in $12 x \leq 120$ by $12$ .

$\frac{12 x}{12} \leq \frac{120}{12}$

Simplify the left side.

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12 x}{12} \leq \frac{120}{12}$

Divide $x$ by $1$ .

$x \leq \frac{120}{12}$

Simplify the right side.

Divide $120$ by $12$ .

$x \leq 10$

Find the domain of $2 \sqrt{3 (x + 2)} - 12$ .

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

$3 (x + 2) \geq 0$

Solve for $x$ .

Divide each term in $3 (x + 2) \geq 0$ by $3$ and simplify.

Divide each term in $3 (x + 2) \geq 0$ by $3$ .

$\frac{3 (x + 2)}{3} \geq \frac{0}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 (x + 2)}{3} \geq \frac{0}{3}$

Divide $x + 2$ by $1$ .

$x + 2 \geq \frac{0}{3}$

Simplify the right side.

Divide $0$ by $3$ .

$x + 2 \geq 0$

Subtract $2$ from both sides of the inequality.

$x \geq - 2$

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

$[- 2, \infty)$

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 10$

$x > 10$

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

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

$x = - 4$

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

$2 \sqrt{3 (- 4) + 6} - 8 \leq 4$

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 $- 2 < x < 10$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 10$ and see if this value makes the original inequality true.

$x = 0$

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

$2 \sqrt{3 (0) + 6} - 8 \leq 4$

The left side $- 3.10102051$ is less than the right side $4$ , which means that the given statement is always true.

True

Test a value on the interval $x > 10$ to see if it makes the inequality true.

Choose a value on the interval $x > 10$ and see if this value makes the original inequality true.

$x = 12$

Replace $x$ with $12$ in the original inequality.

$2 \sqrt{3 (12) + 6} - 8 \leq 4$

The left side $4.96148139$ is greater than the right side $4$ , which means that the given statement is false.

False

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

$x < - 2$ False

$- 2 < x < 10$ True

$x > 10$ False

$x < - 2$ False

$- 2 < x < 10$ True

$x > 10$ False

The solution consists of all of the true intervals.

$- 2 \leq x \leq 10$

The result can be shown in multiple forms.

Inequality Form:

$- 2 \leq x \leq 10$

Interval Notation:

$[- 2, 10]$

Solve for x 2 square root of 3x+6-8<=4

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