Precalculus Examples

$\frac{x - 5}{x^{2} + 25} - 3 x$

Step 1

Set the denominator in $\frac{x - 5}{x^{2} + 25}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 25 = 0$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

Subtract $25$ from both sides of the equation.

$x^{2} = - 25$

Step 2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 25}$

Step 2.3

Simplify $\pm \sqrt{- 25}$ .

Tap for more steps...

Step 2.3.1

Rewrite $- 25$ as $- 1 (25)$ .

$x = \pm \sqrt{- 1 (25)}$

Step 2.3.2

Rewrite $\sqrt{- 1 (25)}$ as $\sqrt{- 1} \cdot \sqrt{25}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{25}$

Step 2.3.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{25}$

Step 2.3.4

Rewrite $25$ as $5^{2}$ .

$x = \pm i \cdot \sqrt{5^{2}}$

Step 2.3.5

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm i \cdot 5$

Step 2.3.6

Move $5$ to the left of $i$ .

$x = \pm 5 i$

Step 2.4

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

Tap for more steps...

Step 2.4.1

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

$x = 5 i$

Step 2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 5 i$

Step 2.4.3

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

$x = 5 i, - 5 i$

Step 3

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 4