Algebra Examples

$x^{2} - 48 > - 5 x + 2$

Step 1

Add $5 x$ to both sides of the inequality.

$x^{2} - 48 + 5 x > 2$

Step 2

Convert the inequality to an equation.

$x^{2} - 48 + 5 x = 2$

Step 3

Subtract $2$ from both sides of the equation.

$x^{2} - 48 + 5 x - 2 = 0$

Step 4

Subtract $2$ from $- 48$ .

$x^{2} + 5 x - 50 = 0$

Step 5

Factor $x^{2} + 5 x - 50$ using the AC method.

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Step 5.1

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 $- 50$ and whose sum is $5$ .

$- 5, 10$

Step 5.2

Write the factored form using these integers.

$(x - 5) (x + 10) = 0$

Step 6

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

$x - 5 = 0$

$x + 10 = 0$

Step 7

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

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Step 7.1

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

$x - 5 = 0$

Step 7.2

Add $5$ to both sides of the equation.

$x = 5$

Step 8

Set $x + 10$ equal to $0$ and solve for $x$ .

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Step 8.1

Set $x + 10$ equal to $0$ .

$x + 10 = 0$

Step 8.2

Subtract $10$ from both sides of the equation.

$x = - 10$

Step 9

The final solution is all the values that make $(x - 5) (x + 10) = 0$ true.

$x = 5, - 10$

Step 10

Use each root to create test intervals.

$x < - 10$

$- 10 < x < 5$

$x > 5$

Step 11

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 11.1

Test a value on the interval $x < - 10$ to see if it makes the inequality true.

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Step 11.1.1

Choose a value on the interval $x < - 10$ and see if this value makes the original inequality true.

$x = - 12$

Step 11.1.2

Replace $x$ with $- 12$ in the original inequality.

${(- 12)}^{2} - 48 > - 5 \cdot - 12 + 2$

Step 11.1.3

The left side $96$ is greater than the right side $62$ , which means that the given statement is always true.

True

Step 11.2

Test a value on the interval $- 10 < x < 5$ to see if it makes the inequality true.

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Step 11.2.1

Choose a value on the interval $- 10 < x < 5$ and see if this value makes the original inequality true.

$x = 0$

Step 11.2.2

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

${(0)}^{2} - 48 > - 5 \cdot 0 + 2$

Step 11.2.3

The left side $- 48$ is not greater than the right side $2$ , which means that the given statement is false.

False

Step 11.3

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

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Step 11.3.1

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

$x = 8$

Step 11.3.2

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

${(8)}^{2} - 48 > - 5 \cdot 8 + 2$

Step 11.3.3

The left side $16$ is greater than the right side $- 38$ , which means that the given statement is always true.

True

Step 11.4

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

$x < - 10$ True

$- 10 < x < 5$ False

$x > 5$ True

$x < - 10$ True

$- 10 < x < 5$ False

$x > 5$ True

Step 12

The solution consists of all of the true intervals.

$x < - 10$ or $x > 5$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$x < - 10 or x > 5$

Interval Notation:

$(- \infty, - 10) \cup (5, \infty)$

Step 14