Graph 7 square root of 3x+9-12<=30

$7 \sqrt{3 x + 9} - 12 \leq 30$

Step 1

Move all terms not containing $7 \sqrt{3 x + 9}$ to the right side of the inequality.

Add $12$ to both sides of the inequality.

$7 \sqrt{3 x + 9} \leq 30 + 12$

Add $30$ and $12$ .

$7 \sqrt{3 x + 9} \leq 42$

Step 2

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

${(7 \sqrt{3 x + 9})}^{2} \leq 42^{2}$

Step 3

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 x + 9}$ as ${(3 x + 9)}^{\frac{1}{2}}$ .

${(7 {(3 x + 9)}^{\frac{1}{2}})}^{2} \leq 42^{2}$

Simplify the left side.

Simplify ${(7 {(3 x + 9)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $7 {(3 x + 9)}^{\frac{1}{2}}$ .

$7^{2} {({(3 x + 9)}^{\frac{1}{2}})}^{2} \leq 42^{2}$

Raise $7$ to the power of $2$ .

$49 {({(3 x + 9)}^{\frac{1}{2}})}^{2} \leq 42^{2}$

Multiply the exponents in ${({(3 x + 9)}^{\frac{1}{2}})}^{2}$ .

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

$49 {(3 x + 9)}^{\frac{1}{2} \cdot 2} \leq 42^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$49 {(3 x + 9)}^{\frac{1}{2} \cdot 2} \leq 42^{2}$

Rewrite the expression.

$49 {(3 x + 9)}^{1} \leq 42^{2}$

Simplify.

$49 (3 x + 9) \leq 42^{2}$

Apply the distributive property.

$49 (3 x) + 49 \cdot 9 \leq 42^{2}$

Multiply.

Multiply $3$ by $49$ .

$147 x + 49 \cdot 9 \leq 42^{2}$

Multiply $49$ by $9$ .

$147 x + 441 \leq 42^{2}$

Simplify the right side.

Raise $42$ to the power of $2$ .

$147 x + 441 \leq 1764$

Step 4

Solve for $x$ .

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

Subtract $441$ from both sides of the inequality.

$147 x \leq 1764 - 441$

Subtract $441$ from $1764$ .

$147 x \leq 1323$

Divide each term in $147 x \leq 1323$ by $147$ and simplify.

Divide each term in $147 x \leq 1323$ by $147$ .

$\frac{147 x}{147} \leq \frac{1323}{147}$

Simplify the left side.

Cancel the common factor of $147$ .

Cancel the common factor.

$\frac{147 x}{147} \leq \frac{1323}{147}$

Divide $x$ by $1$ .

$x \leq \frac{1323}{147}$

Simplify the right side.

Divide $1323$ by $147$ .

$x \leq 9$

Step 5

Find the domain of $7 \sqrt{3 (x + 3)} - 42$ .

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

$3 (x + 3) \geq 0$

Solve for $x$ .

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

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

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x + 3$ by $1$ .

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

Simplify the right side.

Divide $0$ by $3$ .

$x + 3 \geq 0$

Subtract $3$ from both sides of the inequality.

$x \geq - 3$

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

$[- 3, \infty)$

Step 6

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 9$

$x > 9$

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

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

$x = - 6$

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

$7 \sqrt{3 (- 6) + 9} - 12 \leq 30$

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 $- 3 < x < 9$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 9$ and see if this value makes the original inequality true.

$x = 0$

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

$7 \sqrt{3 (0) + 9} - 12 \leq 30$

The left side $9$ is less than the right side $30$ , which means that the given statement is always true.

True

Test a value on the interval $x > 9$ to see if it makes the inequality true.

Choose a value on the interval $x > 9$ and see if this value makes the original inequality true.

$x = 12$

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

$7 \sqrt{3 (12) + 9} - 12 \leq 30$

The left side $34.95742752$ is greater than the right side $30$ , which means that the given statement is false.

False

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

$x < - 3$ False

$- 3 < x < 9$ True

$x > 9$ False

$x < - 3$ False

$- 3 < x < 9$ True

$x > 9$ False

Step 8

The solution consists of all of the true intervals.

$- 3 \leq x \leq 9$

Step 9

Graph 7 square root of 3x+9-12<=30

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