Solve for x square root of 3x-12+7<=16

$\sqrt{3 x - 12} + 7 \leq 16$

Step 1

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

Subtract $7$ from both sides of the inequality.

$\sqrt{3 x - 12} \leq 16 - 7$

Subtract $7$ from $16$ .

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

Step 2

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

${\sqrt{3 x - 12}}^{2} \leq 9^{2}$

Step 3

Simplify each side of the inequality.

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

${({(3 x - 12)}^{\frac{1}{2}})}^{2} \leq 9^{2}$

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, ${(a^{m})}^{n} = a^{m n}$ .

${(3 x - 12)}^{\frac{1}{2} \cdot 2} \leq 9^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(3 x - 12)}^{\frac{1}{2} \cdot 2} \leq 9^{2}$

Rewrite the expression.

${(3 x - 12)}^{1} \leq 9^{2}$

Simplify.

$3 x - 12 \leq 9^{2}$

Simplify the right side.

Raise $9$ to the power of $2$ .

$3 x - 12 \leq 81$

Step 4

Solve for $x$ .

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

Add $12$ to both sides of the inequality.

$3 x \leq 81 + 12$

Add $81$ and $12$ .

$3 x \leq 93$

Divide each term in $3 x \leq 93$ by $3$ and simplify.

Divide each term in $3 x \leq 93$ by $3$ .

$\frac{3 x}{3} \leq \frac{93}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \leq \frac{93}{3}$

Divide $x$ by $1$ .

$x \leq \frac{93}{3}$

Simplify the right side.

Divide $93$ by $3$ .

$x \leq 31$

Step 5

Find the domain of $\sqrt{3 (x - 4)} - 9$ .

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 $3$ .

$x - 4 \geq 0$

Add $4$ to 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.

$x < 4$

$4 < x < 31$

$x > 31$

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.

$x = 0$

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

$\sqrt{3 (0) - 12} + 7 \leq 16$

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 $4 < x < 31$ to see if it makes the inequality true.

Choose a value on the interval $4 < x < 31$ and see if this value makes the original inequality true.

$x = 6$

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

$\sqrt{3 (6) - 12} + 7 \leq 16$

The left side $9.44948974$ is less than the right side $16$ , which means that the given statement is always true.

True

Test a value on the interval $x > 31$ to see if it makes the inequality true.

Choose a value on the interval $x > 31$ and see if this value makes the original inequality true.

$x = 34$

Replace $x$ with $34$ in the original inequality.

$\sqrt{3 (34) - 12} + 7 \leq 16$

The left side $16.48683298$ is greater than the right side $16$ , which means that the given statement is false.

False

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

$x < 4$ False

$4 < x < 31$ True

$x > 31$ False

$x < 4$ False

$4 < x < 31$ True

$x > 31$ False

Step 8

The solution consists of all of the true intervals.

$4 \leq x \leq 31$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$4 \leq x \leq 31$

Interval Notation:

$[4, 31]$

Step 10

Solve for x square root of 3x-12+7<=16

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