Finite Math Examples

$| x^{2} - 5 | < 4 x$

Step 1

Write $| x^{2} - 5 | < 4 x$ as a piecewise.

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Step 1.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x^{2} - 5 \geq 0$

Step 1.2

Solve the inequality.

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Step 1.2.1

Add $5$ to both sides of the inequality.

$x^{2} \geq 5$

Step 1.2.2

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

$\sqrt{x^{2}} \geq \sqrt{5}$

Step 1.2.3

Simplify the left side.

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Step 1.2.3.1

Pull terms out from under the radical.

$| x | \geq \sqrt{5}$

Step 1.2.4

Write $| x | \geq \sqrt{5}$ as a piecewise.

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Step 1.2.4.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 1.2.4.2

In the piece where $x$ is non-negative, remove the absolute value.

$x \geq \sqrt{5}$

Step 1.2.4.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 1.2.4.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \geq \sqrt{5}$

Step 1.2.4.5

Write as a piecewise.

${\begin{cases} x \geq \sqrt{5} & x \geq 0 \\ - x \geq \sqrt{5} & x < 0 \end{cases}$

Step 1.2.5

Find the intersection of $x \geq \sqrt{5}$ and $x \geq 0$ .

$x \geq \sqrt{5}$

Step 1.2.6

Divide each term in $- x \geq \sqrt{5}$ by $- 1$ and simplify.

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Step 1.2.6.1

Divide each term in $- x \geq \sqrt{5}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{\sqrt{5}}{- 1}$

Step 1.2.6.2

Simplify the left side.

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Step 1.2.6.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{\sqrt{5}}{- 1}$

Step 1.2.6.2.2

Divide $x$ by $1$ .

$x \leq \frac{\sqrt{5}}{- 1}$

Step 1.2.6.3

Simplify the right side.

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Step 1.2.6.3.1

Move the negative one from the denominator of $\frac{\sqrt{5}}{- 1}$ .

$x \leq - 1 \cdot \sqrt{5}$

Step 1.2.6.3.2

Rewrite $- 1 \cdot \sqrt{5}$ as $- \sqrt{5}$ .

$x \leq - \sqrt{5}$

Step 1.2.7

Find the union of the solutions.

$x \leq - \sqrt{5}$ or $x \geq \sqrt{5}$

Step 1.3

In the piece where $x^{2} - 5$ is non-negative, remove the absolute value.

$x^{2} - 5 < 4 x$

Step 1.4

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x^{2} - 5 < 0$

Step 1.5

Solve the inequality.

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Step 1.5.1

Add $5$ to both sides of the inequality.

$x^{2} < 5$

Step 1.5.2

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

$\sqrt{x^{2}} < \sqrt{5}$

Step 1.5.3

Simplify the left side.

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Step 1.5.3.1

Pull terms out from under the radical.

$| x | < \sqrt{5}$

Step 1.5.4

Write $| x | < \sqrt{5}$ as a piecewise.

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Step 1.5.4.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 1.5.4.2

In the piece where $x$ is non-negative, remove the absolute value.

$x < \sqrt{5}$

Step 1.5.4.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 1.5.4.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x < \sqrt{5}$

Step 1.5.4.5

Write as a piecewise.

${\begin{cases} x < \sqrt{5} & x \geq 0 \\ - x < \sqrt{5} & x < 0 \end{cases}$

Step 1.5.5

Find the intersection of $x < \sqrt{5}$ and $x \geq 0$ .

$0 \leq x < \sqrt{5}$

Step 1.5.6

Solve $- x < \sqrt{5}$ when $x < 0$ .

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Step 1.5.6.1

Divide each term in $- x < \sqrt{5}$ by $- 1$ and simplify.

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Step 1.5.6.1.1

Divide each term in $- x < \sqrt{5}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} > \frac{\sqrt{5}}{- 1}$

Step 1.5.6.1.2

Simplify the left side.

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Step 1.5.6.1.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} > \frac{\sqrt{5}}{- 1}$

Step 1.5.6.1.2.2

Divide $x$ by $1$ .

$x > \frac{\sqrt{5}}{- 1}$

Step 1.5.6.1.3

Simplify the right side.

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Step 1.5.6.1.3.1

Move the negative one from the denominator of $\frac{\sqrt{5}}{- 1}$ .

$x > - 1 \cdot \sqrt{5}$

Step 1.5.6.1.3.2

Rewrite $- 1 \cdot \sqrt{5}$ as $- \sqrt{5}$ .

$x > - \sqrt{5}$

Step 1.5.6.2

Find the intersection of $x > - \sqrt{5}$ and $x < 0$ .

$- \sqrt{5} < x < 0$

Step 1.5.7

Find the union of the solutions.

$- \sqrt{5} < x < \sqrt{5}$

Step 1.6

In the piece where $x^{2} - 5$ is negative, remove the absolute value and multiply by $- 1$ .

$- (x^{2} - 5) < 4 x$

Step 1.7

Write as a piecewise.

${\begin{cases} x^{2} - 5 < 4 x & x \leq - \sqrt{5} or x \geq \sqrt{5} \\ - (x^{2} - 5) < 4 x & - \sqrt{5} < x < \sqrt{5} \end{cases}$

Step 1.8

Simplify $- (x^{2} - 5) < 4 x$ .

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Step 1.8.1

Apply the distributive property.

${\begin{cases} x^{2} - 5 < 4 x & x \leq - \sqrt{5} or x \geq \sqrt{5} \\ - x^{2} - - 5 < 4 x & - \sqrt{5} < x < \sqrt{5} \end{cases}$

Step 1.8.2

Multiply $- 1$ by $- 5$ .

${\begin{cases} x^{2} - 5 < 4 x & x \leq - \sqrt{5} or x \geq \sqrt{5} \\ - x^{2} + 5 < 4 x & - \sqrt{5} < x < \sqrt{5} \end{cases}$

Step 2

Solve $x^{2} - 5 < 4 x$ when $x \leq - \sqrt{5} or x \geq \sqrt{5}$ .

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Step 2.1

Solve $x^{2} - 5 < 4 x$ for $x$ .

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Step 2.1.1

Subtract $4 x$ from both sides of the inequality.

$x^{2} - 5 - 4 x < 0$

Step 2.1.2

Convert the inequality to an equation.

$x^{2} - 5 - 4 x = 0$

Step 2.1.3

Factor $x^{2} - 5 - 4 x$ using the AC method.

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Step 2.1.3.1

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 $- 5$ and whose sum is $- 4$ .

$- 5, 1$

Step 2.1.3.2

Write the factored form using these integers.

$(x - 5) (x + 1) = 0$

Step 2.1.4

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

$x - 5 = 0$

$x + 1 = 0$

Step 2.1.5

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

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Step 2.1.5.1

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

$x - 5 = 0$

Step 2.1.5.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.1.6

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

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Step 2.1.6.1

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

$x + 1 = 0$

Step 2.1.6.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.1.7

The final solution is all the values that make $(x - 5) (x + 1) = 0$ true.

$x = 5, - 1$

Step 2.1.8

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 5$

$x > 5$

Step 2.1.9

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 2.1.9.1

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

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Step 2.1.9.1.1

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

$x = - 4$

Step 2.1.9.1.2

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

${(- 4)}^{2} - 5 < 4 (- 4)$

Step 2.1.9.1.3

The left side $11$ is not less than the right side $- 16$ , which means that the given statement is false.

False

Step 2.1.9.2

Test a value on the interval $- 1 < x < 5$ to see if it makes the inequality true.

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Step 2.1.9.2.1

Choose a value on the interval $- 1 < x < 5$ and see if this value makes the original inequality true.

$x = 0$

Step 2.1.9.2.2

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

${(0)}^{2} - 5 < 4 (0)$

Step 2.1.9.2.3

The left side $- 5$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 2.1.9.3

Test a value on the interval $x > 5$ to see if it makes the inequality true.

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Step 2.1.9.3.1

Choose a value on the interval $x > 5$ and see if this value makes the original inequality true.

$x = 8$

Step 2.1.9.3.2

Replace $x$ with $8$ in the original inequality.

${(8)}^{2} - 5 < 4 (8)$

Step 2.1.9.3.3

The left side $59$ is not less than the right side $32$ , which means that the given statement is false.

False

Step 2.1.9.4

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

$x < - 1$ False

$- 1 < x < 5$ True

$x > 5$ False

$x < - 1$ False

$- 1 < x < 5$ True

$x > 5$ False

Step 2.1.10

The solution consists of all of the true intervals.

$- 1 < x < 5$

Step 2.2

Find the intersection of $- 1 < x < 5$ and $x \leq - \sqrt{5} o r + x \geq \sqrt{5}$ .

$\sqrt{5} \leq x < 5$

Step 3

Solve $- x^{2} + 5 < 4 x$ when $- \sqrt{5} < x < \sqrt{5}$ .

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Step 3.1

Solve $- x^{2} + 5 < 4 x$ for $x$ .

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Step 3.1.1

Subtract $4 x$ from both sides of the inequality.

$- x^{2} + 5 - 4 x < 0$

Step 3.1.2

Convert the inequality to an equation.

$- x^{2} + 5 - 4 x = 0$

Step 3.1.3

Factor the left side of the equation.

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Step 3.1.3.1

Factor $- 1$ out of $- x^{2} + 5 - 4 x$ .

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Step 3.1.3.1.1

Move $5$ .

$- x^{2} - 4 x + 5 = 0$

Step 3.1.3.1.2

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

$- (x^{2}) - 4 x + 5 = 0$

Step 3.1.3.1.3

Factor $- 1$ out of $- 4 x$ .

$- (x^{2}) - (4 x) + 5 = 0$

Step 3.1.3.1.4

Rewrite $5$ as $- 1 (- 5)$ .

$- (x^{2}) - (4 x) - 1 \cdot - 5 = 0$

Step 3.1.3.1.5

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

$- (x^{2} + 4 x) - 1 \cdot - 5 = 0$

Step 3.1.3.1.6

Factor $- 1$ out of $- (x^{2} + 4 x) - 1 (- 5)$ .

$- (x^{2} + 4 x - 5) = 0$

Step 3.1.3.2

Factor.

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Step 3.1.3.2.1

Factor $x^{2} + 4 x - 5$ using the AC method.

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Step 3.1.3.2.1.1

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 $- 5$ and whose sum is $4$ .

$- 1, 5$

Step 3.1.3.2.1.2

Write the factored form using these integers.

$- ((x - 1) (x + 5)) = 0$

Step 3.1.3.2.2

Remove unnecessary parentheses.

$- (x - 1) (x + 5) = 0$

Step 3.1.4

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

$x - 1 = 0$

$x + 5 = 0$

Step 3.1.5

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

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Step 3.1.5.1

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

$x - 1 = 0$

Step 3.1.5.2

Add $1$ to both sides of the equation.

$x = 1$

Step 3.1.6

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

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Step 3.1.6.1

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

$x + 5 = 0$

Step 3.1.6.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 3.1.7

The final solution is all the values that make $- (x - 1) (x + 5) = 0$ true.

$x = 1, - 5$

Step 3.1.8

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 1$

$x > 1$

Step 3.1.9

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 3.1.9.1

Test a value on the interval $x < - 5$ to see if it makes the inequality true.

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Step 3.1.9.1.1

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Step 3.1.9.1.2

Replace $x$ with $- 8$ in the original inequality.

$- {(- 8)}^{2} + 5 < 4 (- 8)$

Step 3.1.9.1.3

The left side $- 59$ is less than the right side $- 32$ , which means that the given statement is always true.

True

Step 3.1.9.2

Test a value on the interval $- 5 < x < 1$ to see if it makes the inequality true.

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Step 3.1.9.2.1

Choose a value on the interval $- 5 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 3.1.9.2.2

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

$- {(0)}^{2} + 5 < 4 (0)$

Step 3.1.9.2.3

The left side $5$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 3.1.9.3

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

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Step 3.1.9.3.1

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

$x = 4$

Step 3.1.9.3.2

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

$- {(4)}^{2} + 5 < 4 (4)$

Step 3.1.9.3.3

The left side $- 11$ is less than the right side $16$ , which means that the given statement is always true.

True

Step 3.1.9.4

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

$x < - 5$ True

$- 5 < x < 1$ False

$x > 1$ True

$x < - 5$ True

$- 5 < x < 1$ False

$x > 1$ True

Step 3.1.10

The solution consists of all of the true intervals.

$x < - 5$ or $x > 1$

Step 3.2

Find the intersection of $x < - 5 or x > 1$ and $- \sqrt{5} < x < \sqrt{5}$ .

$1 < x < \sqrt{5}$

Step 4

Find the union of the solutions.

$1 < x < 5$

Step 5

Convert the inequality to interval notation.

$(1, 5)$

Step 6