Solve the Inequality for x square root of x<3

$\sqrt{x} < 3$

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

${\sqrt{x}}^{2} < 3^{2}$

Simplify each side of the inequality.

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

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

Simplify the left side.

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

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

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

$x^{\frac{1}{2} \cdot 2} < 3^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x^{\frac{1}{2} \cdot 2} < 3^{2}$

Rewrite the expression.

$x^{1} < 3^{2}$

Simplify.

$x < 3^{2}$

Simplify the right side.

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

$x < 9$

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

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.

$x < 0$

$0 < x < 9$

$x > 9$

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.

$\sqrt{- 2} < 3$

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 $0 < x < 9$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 9$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

$\sqrt{4} < 3$

The left side $2$ is less than the right side $3$ , 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.

$\sqrt{12} < 3$

The left side $3.46410161$ is not less than the right side $3$ , which means that the given statement is false.

False

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

$x < 0$ False

$0 < x < 9$ True

$x > 9$ False

$x < 0$ False

$0 < x < 9$ True

$x > 9$ False

The solution consists of all of the true intervals.

$0 < x < 9$

The result can be shown in multiple forms.

Inequality Form:

$0 < x < 9$

Interval Notation:

$(0, 9)$

Solve the Inequality for x square root of x<3

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