Algebra Examples

$\frac{x^{2} + 16 x - 57}{36 - 4 x^{2}}$

Step 1

Set the denominator in $\frac{x^{2} + 16 x - 57}{36 - 4 x^{2}}$ equal to $0$ to find where the expression is undefined.

$36 - 4 x^{2} = 0$

Step 2

Solve for $x$ .

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Step 2.1

Subtract $36$ from both sides of the equation.

$- 4 x^{2} = - 36$

Step 2.2

Divide each term in $- 4 x^{2} = - 36$ by $- 4$ and simplify.

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Step 2.2.1

Divide each term in $- 4 x^{2} = - 36$ by $- 4$ .

$\frac{- 4 x^{2}}{- 4} = \frac{- 36}{- 4}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $- 4$ .

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Step 2.2.2.1.1

Cancel the common factor.

$\frac{- 4 x^{2}}{- 4} = \frac{- 36}{- 4}$

Step 2.2.2.1.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 36}{- 4}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $- 36$ by $- 4$ .

$x^{2} = 9$

Step 2.3

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

$x = \pm \sqrt{9}$

Step 2.4

Simplify $\pm \sqrt{9}$ .

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Step 2.4.1

Rewrite $9$ as $3^{2}$ .

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

Step 2.4.2

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

$x = \pm 3$

Step 2.5

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

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Step 2.5.1

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

$x = 3$

Step 2.5.2

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

$x = - 3$

Step 2.5.3

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

$x = 3, - 3$

Step 3

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = - 3, x = 3$

Step 4