Algebra Examples

x^{2} - 10 x \leq - 9

$x^{2} - 10 x \leq - 9$

Step 1

Convert the inequality to an equation.

x^{2} - 10 x = - 9

$x^{2} - 10 x = - 9$

Step 2

Add

9

$9$ to both sides of the equation.

x^{2} - 10 x + 9 = 0

$x^{2} - 10 x + 9 = 0$

Step 3

Factor

x^{2} - 10 x + 9

$x^{2} - 10 x + 9$ using the AC method.

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Step 3.1

Consider the form

x^{2} + b x + c

$x^{2} + b x + c$ . Find a pair of integers whose product is

c

$c$ and whose sum is

b

$b$ . In this case, whose product is

9

$9$ and whose sum is

- 10

$- 10$ .

- 9, - 1

$- 9, - 1$

Step 3.2

Write the factored form using these integers.

(x - 9) (x - 1) = 0

$(x - 9) (x - 1) = 0$

(x - 9) (x - 1) = 0

$(x - 9) (x - 1) = 0$

Step 4

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

x - 9 = 0

$x - 9 = 0$

x - 1 = 0

$x - 1 = 0$

Step 5

Set

x - 9

$x - 9$ equal to

0

$0$ and solve for

x

$x$ .

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Step 5.1

Set

x - 9

$x - 9$ equal to

0

$0$ .

x - 9 = 0

$x - 9 = 0$

Step 5.2

Add

9

$9$ to both sides of the equation.

x = 9

$x = 9$

x = 9

$x = 9$

Step 6

Set

x - 1

$x - 1$ equal to

0

$0$ and solve for

x

$x$ .

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Step 6.1

Set

x - 1

$x - 1$ equal to

0

$0$ .

x - 1 = 0

$x - 1 = 0$

Step 6.2

Add

1

$1$ to both sides of the equation.

x = 1

$x = 1$

x = 1

$x = 1$

Step 7

The final solution is all the values that make

(x - 9) (x - 1) = 0

$(x - 9) (x - 1) = 0$ true.

x = 9, 1

$x = 9, 1$

Step 8

Use each root to create test intervals.

x < 1

$x < 1$

1 < x < 9

$1 < x < 9$

x > 9

$x > 9$

Step 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 9.1

Test a value on the interval

x < 1

$x < 1$ to see if it makes the inequality true.

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Step 9.1.1

Choose a value on the interval

x < 1

$x < 1$ and see if this value makes the original inequality true.

x = 0

$x = 0$

Step 9.1.2

Replace

x

$x$ with

0

$0$ in the original inequality.

{(0)}^{2} - 10 \cdot 0 \leq - 9

${(0)}^{2} - 10 \cdot 0 \leq - 9$

Step 9.1.3

The left side

0

$0$ is greater than the right side

- 9

$- 9$ , which means that the given statement is false.

False

Step 9.2

Test a value on the interval

1 < x < 9

$1 < x < 9$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval

1 < x < 9

$1 < x < 9$ and see if this value makes the original inequality true.

x = 5

$x = 5$

Step 9.2.2

Replace

x

$x$ with

5

$5$ in the original inequality.

{(5)}^{2} - 10 \cdot 5 \leq - 9

${(5)}^{2} - 10 \cdot 5 \leq - 9$

Step 9.2.3

The left side

- 25

$- 25$ is less than the right side

- 9

$- 9$ , which means that the given statement is always true.

True

Step 9.3

Test a value on the interval

x > 9

$x > 9$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval

x > 9

$x > 9$ and see if this value makes the original inequality true.

x = 12

$x = 12$

Step 9.3.2

Replace

x

$x$ with

12

$12$ in the original inequality.

{(12)}^{2} - 10 \cdot 12 \leq - 9

${(12)}^{2} - 10 \cdot 12 \leq - 9$

Step 9.3.3

The left side

24

$24$ is greater than the right side

- 9

$- 9$ , which means that the given statement is false.

False

Step 9.4

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

x < 1

$x < 1$ False

1 < x < 9

$1 < x < 9$ True

x > 9

$x > 9$ False

x < 1

$x < 1$ False

1 < x < 9

$1 < x < 9$ True

x > 9

$x > 9$ False

Step 10

The solution consists of all of the true intervals.

1 \leq x \leq 9

$1 \leq x \leq 9$

Step 11

The result can be shown in multiple forms.

Inequality Form:

1 \leq x \leq 9

$1 \leq x \leq 9$

Interval Notation:

[1, 9]

$[1, 9]$

Step 12

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