Solve the Inequality for x 3 square root of x<=12

$3 \sqrt{x} \leq 12$

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

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

Simplify each side of the inequality.

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

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

Simplify the left side.

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

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

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

Raise $3$ to the power of $2$ .

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

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

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

$9 x^{\frac{1}{2} \cdot 2} \leq 12^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$9 x^{\frac{1}{2} \cdot 2} \leq 12^{2}$

Rewrite the expression.

$9 x^{1} \leq 12^{2}$

Simplify.

$9 x \leq 12^{2}$

Simplify the right side.

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

$9 x \leq 144$

Divide each term in $9 x \leq 144$ by $9$ and simplify.

Divide each term in $9 x \leq 144$ by $9$ .

$\frac{9 x}{9} \leq \frac{144}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 x}{9} \leq \frac{144}{9}$

Divide $x$ by $1$ .

$x \leq \frac{144}{9}$

Simplify the right side.

Divide $144$ by $9$ .

$x \leq 16$

Find the domain of $3 \sqrt{x} - 12$ .

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

$x \geq 0$

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

$[0, \infty)$

Use each root to create test intervals.

$x < 0$

$0 < x < 16$

$x > 16$

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 $x < 0$ to see if it makes the inequality true.

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Replace $x$ with $- 2$ in the original inequality.

$3 \sqrt{- 2} \leq 12$

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 $0 < x < 16$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 16$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

$3 \sqrt{2} \leq 12$

The left side $4.24264068$ is less than the right side $12$ , which means that the given statement is always true.

True

Test a value on the interval $x > 16$ to see if it makes the inequality true.

Choose a value on the interval $x > 16$ and see if this value makes the original inequality true.

$x = 18$

Replace $x$ with $18$ in the original inequality.

$3 \sqrt{18} \leq 12$

The left side $12.72792206$ is greater than the right side $12$ , which means that the given statement is false.

False

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

$x < 0$ False

$0 < x < 16$ True

$x > 16$ False

$x < 0$ False

$0 < x < 16$ True

$x > 16$ False

The solution consists of all of the true intervals.

$0 \leq x \leq 16$

The result can be shown in multiple forms.

Inequality Form:

$0 \leq x \leq 16$

Interval Notation:

$[0, 16]$

Solve the Inequality for x 3 square root of x<=12

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