Finite Math Examples

$\frac{x^{2} - 11 x + 30}{x^{2} - x - 20} \cdot \frac{x^{2} - 16}{x^{2} - 36}$

Step 1

Set the denominator in $\frac{x^{2} - 11 x + 30}{x^{2} - x - 20}$ equal to $0$ to find where the expression is undefined.

$x^{2} - x - 20 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor $x^{2} - x - 20$ using the AC method.

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Step 2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 20$ and whose sum is $- 1$ .

$- 5, 4$

Step 2.1.2

Write the factored form using these integers.

$(x - 5) (x + 4) = 0$

Step 2.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 5 = 0$

$x + 4 = 0$

Step 2.3

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 2.3.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.4

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 2.4.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 2.5

The final solution is all the values that make $(x - 5) (x + 4) = 0$ true.

$x = 5, - 4$

Step 3

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

$x^{2} - 36 = 0$

Step 4

Solve for $x$ .

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Step 4.1

Add $36$ to both sides of the equation.

$x^{2} = 36$

Step 4.2

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

$x = \pm \sqrt{36}$

Step 4.3

Simplify $\pm \sqrt{36}$ .

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Step 4.3.1

Rewrite $36$ as $6^{2}$ .

$x = \pm \sqrt{6^{2}}$

Step 4.3.2

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

$x = \pm 6$

Step 4.4

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

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Step 4.4.1

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

$x = 6$

Step 4.4.2

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

$x = - 6$

Step 4.4.3

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

$x = 6, - 6$

Step 5

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 $0$ .

$x = - 6, x = - 4, x = 5, x = 6$

Step 6