Solve for x 6 square root of 8x-32-13<=11

$6 \sqrt{8 x - 32} - 13 \leq 11$

Step 1

Move all terms not containing $6 \sqrt{8 x - 32}$ to the right side of the inequality.

Add $13$ to both sides of the inequality.

$6 \sqrt{8 x - 32} \leq 11 + 13$

Add $11$ and $13$ .

$6 \sqrt{8 x - 32} \leq 24$

Step 2

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

${(6 \sqrt{8 x - 32})}^{2} \leq 24^{2}$

Step 3

Simplify each side of the inequality.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{8 x - 32}$ as ${(8 x - 32)}^{\frac{1}{2}}$ .

${(6 {(8 x - 32)}^{\frac{1}{2}})}^{2} \leq 24^{2}$

Simplify the left side.

Simplify ${(6 {(8 x - 32)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $6 {(8 x - 32)}^{\frac{1}{2}}$ .

$6^{2} {({(8 x - 32)}^{\frac{1}{2}})}^{2} \leq 24^{2}$

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

$36 {({(8 x - 32)}^{\frac{1}{2}})}^{2} \leq 24^{2}$

Multiply the exponents in ${({(8 x - 32)}^{\frac{1}{2}})}^{2}$ .

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

$36 {(8 x - 32)}^{\frac{1}{2} \cdot 2} \leq 24^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$36 {(8 x - 32)}^{\frac{1}{2} \cdot 2} \leq 24^{2}$

Rewrite the expression.

$36 {(8 x - 32)}^{1} \leq 24^{2}$

Simplify.

$36 (8 x - 32) \leq 24^{2}$

Apply the distributive property.

$36 (8 x) + 36 \cdot - 32 \leq 24^{2}$

Multiply.

Multiply $8$ by $36$ .

$288 x + 36 \cdot - 32 \leq 24^{2}$

Multiply $36$ by $- 32$ .

$288 x - 1152 \leq 24^{2}$

Simplify the right side.

Raise $24$ to the power of $2$ .

$288 x - 1152 \leq 576$

Step 4

Solve for $x$ .

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

Add $1152$ to both sides of the inequality.

$288 x \leq 576 + 1152$

Add $576$ and $1152$ .

$288 x \leq 1728$

Divide each term in $288 x \leq 1728$ by $288$ and simplify.

Divide each term in $288 x \leq 1728$ by $288$ .

$\frac{288 x}{288} \leq \frac{1728}{288}$

Simplify the left side.

Cancel the common factor of $288$ .

Cancel the common factor.

$\frac{288 x}{288} \leq \frac{1728}{288}$

Divide $x$ by $1$ .

$x \leq \frac{1728}{288}$

Simplify the right side.

Divide $1728$ by $288$ .

$x \leq 6$

Step 5

Find the domain of $12 \sqrt{2 (x - 4)} - 24$ .

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

$2 (x - 4) \geq 0$

Solve for $x$ .

Divide each term in $2 (x - 4) \geq 0$ by $2$ and simplify.

Divide each term in $2 (x - 4) \geq 0$ by $2$ .

$\frac{2 (x - 4)}{2} \geq \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 (x - 4)}{2} \geq \frac{0}{2}$

Divide $x - 4$ by $1$ .

$x - 4 \geq \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$x - 4 \geq 0$

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.

$x < 4$

$4 < x < 6$

$x > 6$

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.

$6 \sqrt{8 (0) - 32} - 13 \leq 11$

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 $4 < x < 6$ to see if it makes the inequality true.

Choose a value on the interval $4 < x < 6$ and see if this value makes the original inequality true.

$x = 5$

Replace $x$ with $5$ in the original inequality.

$6 \sqrt{8 (5) - 32} - 13 \leq 11$

The left side $3.97056274$ is less than the right side $11$ , which means that the given statement is always true.

True

Test a value on the interval $x > 6$ to see if it makes the inequality true.

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Replace $x$ with $8$ in the original inequality.

$6 \sqrt{8 (8) - 32} - 13 \leq 11$

The left side $20.94112549$ is greater than the right side $11$ , which means that the given statement is false.

False

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

$x < 4$ False

$4 < x < 6$ True

$x > 6$ False

$x < 4$ False

$4 < x < 6$ True

$x > 6$ False

Step 8

The solution consists of all of the true intervals.

$4 \leq x \leq 6$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$4 \leq x \leq 6$

Interval Notation:

$[4, 6]$

Step 10

Solve for x 6 square root of 8x-32-13<=11

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