Finite Math Examples

$\sqrt[6]{4 - x^{2} - 3 x}$

Step 1

Set the radicand in $\sqrt[6]{4 - x^{2} - 3 x}$ less than $0$ to find where the expression is undefined.

$4 - x^{2} - 3 x < 0$

Step 2

Solve for $x$ .

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Step 2.1

Convert the inequality to an equation.

$4 - x^{2} - 3 x = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Factor $- 1$ out of $4 - x^{2} - 3 x$ .

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Step 2.2.1.1

Move $4$ .

$- x^{2} - 3 x + 4 = 0$

Step 2.2.1.2

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) - 3 x + 4 = 0$

Step 2.2.1.3

Factor $- 1$ out of $- 3 x$ .

$- (x^{2}) - (3 x) + 4 = 0$

Step 2.2.1.4

Rewrite $4$ as $- 1 (- 4)$ .

$- (x^{2}) - (3 x) - 1 \cdot - 4 = 0$

Step 2.2.1.5

Factor $- 1$ out of $- (x^{2}) - (3 x)$ .

$- (x^{2} + 3 x) - 1 \cdot - 4 = 0$

Step 2.2.1.6

Factor $- 1$ out of $- (x^{2} + 3 x) - 1 (- 4)$ .

$- (x^{2} + 3 x - 4) = 0$

Step 2.2.2

Factor.

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Step 2.2.2.1

Factor $x^{2} + 3 x - 4$ using the AC method.

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Step 2.2.2.1.1

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 $- 4$ and whose sum is $3$ .

$- 1, 4$

Step 2.2.2.1.2

Write the factored form using these integers.

$- ((x - 1) (x + 4)) = 0$

Step 2.2.2.2

Remove unnecessary parentheses.

$- (x - 1) (x + 4) = 0$

Step 2.3

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

$x - 1 = 0$

$x + 4 = 0$

Step 2.4

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 2.4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.5

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

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Step 2.5.1

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

$x + 4 = 0$

Step 2.5.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 2.6

The final solution is all the values that make $- (x - 1) (x + 4) = 0$ true.

$x = 1, - 4$

Step 2.7

Use each root to create test intervals.

$x < - 4$

$- 4 < x < 1$

$x > 1$

Step 2.8

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 2.8.1

Test a value on the interval $x < - 4$ to see if it makes the inequality true.

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Step 2.8.1.1

Choose a value on the interval $x < - 4$ and see if this value makes the original inequality true.

$x = - 6$

Step 2.8.1.2

Replace $x$ with $- 6$ in the original inequality.

$4 - {(- 6)}^{2} - 3 \cdot - 6 < 0$

Step 2.8.1.3

The left side $- 14$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 2.8.2

Test a value on the interval $- 4 < x < 1$ to see if it makes the inequality true.

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Step 2.8.2.1

Choose a value on the interval $- 4 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 2.8.2.2

Replace $x$ with $0$ in the original inequality.

$4 - {(0)}^{2} - 3 \cdot 0 < 0$

Step 2.8.2.3

The left side $4$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 2.8.3

Test a value on the interval $x > 1$ to see if it makes the inequality true.

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Step 2.8.3.1

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Step 2.8.3.2

Replace $x$ with $4$ in the original inequality.

$4 - {(4)}^{2} - 3 \cdot 4 < 0$

Step 2.8.3.3

The left side $- 24$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 2.8.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 4$ True

$- 4 < x < 1$ False

$x > 1$ True

$x < - 4$ True

$- 4 < x < 1$ False

$x > 1$ True

Step 2.9

The solution consists of all of the true intervals.

$x < - 4$ or $x > 1$

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 $0$ .

$x < - 4, x > 1$

$(- \infty, - 4) \cup (1, \infty)$

Step 4