Solve the Inequality for x square root of 2x-6<=6

$\sqrt{2 x - 6} \leq 6$

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

${\sqrt{2 x - 6}}^{2} \leq 6^{2}$

Simplify each side of the inequality.

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

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

Simplify the left side.

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

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

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

${(2 x - 6)}^{\frac{1}{2} \cdot 2} \leq 6^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(2 x - 6)}^{\frac{1}{2} \cdot 2} \leq 6^{2}$

Rewrite the expression.

${(2 x - 6)}^{1} \leq 6^{2}$

Simplify.

$2 x - 6 \leq 6^{2}$

Simplify the right side.

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

$2 x - 6 \leq 36$

Solve for $x$ .

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

Add $6$ to both sides of the inequality.

$2 x \leq 36 + 6$

Add $36$ and $6$ .

$2 x \leq 42$

Divide each term in $2 x \leq 42$ by $2$ and simplify.

Divide each term in $2 x \leq 42$ by $2$ .

$\frac{2 x}{2} \leq \frac{42}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \leq \frac{42}{2}$

Divide $x$ by $1$ .

$x \leq \frac{42}{2}$

Simplify the right side.

Divide $42$ by $2$ .

$x \leq 21$

Find the domain of $\sqrt{2 (x - 3)} - 6$ .

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

$2 (x - 3) \geq 0$

Solve for $x$ .

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x - 3$ by $1$ .

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

Simplify the right side.

Divide $0$ by $2$ .

$x - 3 \geq 0$

Add $3$ to both sides of the inequality.

$x \geq 3$

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

$[3, \infty)$

Use each root to create test intervals.

$x < 3$

$3 < x < 21$

$x > 21$

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.

$x = 0$

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

$\sqrt{2 (0) - 6} \leq 6$

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 $3 < x < 21$ to see if it makes the inequality true.

Choose a value on the interval $3 < x < 21$ and see if this value makes the original inequality true.

$x = 6$

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

$\sqrt{2 (6) - 6} \leq 6$

The left side $2.44948974$ is less than the right side $6$ , which means that the given statement is always true.

True

Test a value on the interval $x > 21$ to see if it makes the inequality true.

Choose a value on the interval $x > 21$ and see if this value makes the original inequality true.

$x = 24$

Replace $x$ with $24$ in the original inequality.

$\sqrt{2 (24) - 6} \leq 6$

The left side $6.48074069$ is greater than the right side $6$ , which means that the given statement is false.

False

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

$x < 3$ False

$3 < x < 21$ True

$x > 21$ False

$x < 3$ False

$3 < x < 21$ True

$x > 21$ False

The solution consists of all of the true intervals.

$3 \leq x \leq 21$

The result can be shown in multiple forms.

Inequality Form:

$3 \leq x \leq 21$

Interval Notation:

$[3, 21]$

Solve the Inequality for x square root of 2x-6<=6

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