Find the Domain square root of x to the power of two minus nine

$\sqrt{x^{2} - 9}$

Set the radicand in $\sqrt{x^{2} - 9}$ greater than or equal to $0$ to find where the expression is defined.

$x^{2} - 9 \geq 0$

Calculate $x$ .

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Add $9$ to both sides of the inequality.

$x^{2} \geq 9$

Take the square root of both sides of the inequality to eliminate the exponent on the left side.

$x \geq \pm \sqrt{9}$

The total solution is the sum of the solution's positive and negative components.

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Clarify the right side of the equation.

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Should be rewritten $9$ as $3^{2}$ .

$x \geq \pm \sqrt{3^{2}}$

Withdraw terms from under the radical, assuming positive real numbers.

$x \geq \pm 3$

The complete solution is the result of both the positive and negative portions of the solution.

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First, use the positive value of the $\pm$ to find the Initially solution.

$x \geq 3$

Next, use the negative value of the $\pm$ To calculate the second answer. Since this is an inequality, flip the direction of the inequality sign on the $-$ portion of the solution.

$x \leq - 3$

The complete solution is the result of both the positive and negative portions of the solution.

$x \geq 3$ or $x \leq - 3$

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

Interval Notation:

$(- \infty, - 3] \cup [3, \infty)$

Set-Builder Notation:

${x | x \leq - 3, x \geq 3}$

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