Find the Domain y=( square root of 2x-9)/(2x+9)

$y = \frac{\sqrt{2 x - 9}}{2 x + 9}$

Step 1

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

$2 x - 9 \geq 0$

Step 2

Solve for $x$ .

Add $9$ to both sides of the inequality.

$2 x \geq 9$

Divide each term in $2 x \geq 9$ by $2$ and simplify.

Divide each term in $2 x \geq 9$ by $2$ .

$\frac{2 x}{2} \geq \frac{9}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \geq \frac{9}{2}$

Divide $x$ by $1$ .

$x \geq \frac{9}{2}$

Step 3

Set the denominator in $\frac{\sqrt{2 x - 9}}{2 x + 9}$ equal to $0$ to find where the expression is undefined.

$2 x + 9 = 0$

Step 4

Solve for $x$ .

Subtract $9$ from both sides of the equation.

$2 x = - 9$

Divide each term in $2 x = - 9$ by $2$ and simplify.

Divide each term in $2 x = - 9$ by $2$ .

$\frac{2 x}{2} = \frac{- 9}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 9}{2}$

Divide $x$ by $1$ .

$x = \frac{- 9}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{9}{2}$

Step 5

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

Interval Notation:

$[\frac{9}{2}, \infty)$

Set-Builder Notation:

${x | x \geq \frac{9}{2}}$

Step 6