Algebra Examples

8 < x (7 - x)

$8 < x (7 - x)$

Step 1

Rewrite so

x

$x$ is on the left side of the inequality.

x (7 - x) > 8

$x (7 - x) > 8$

Step 2

Simplify

x (7 - x)

$x (7 - x)$ .

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Step 2.1

Simplify by multiplying through.

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Step 2.1.1

Apply the distributive property.

x \cdot 7 + x (- x) > 8

$x \cdot 7 + x (- x) > 8$

Step 2.1.2

Reorder.

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Step 2.1.2.1

Move

7

$7$ to the left of

x

$x$ .

7 \cdot x + x (- x) > 8

$7 \cdot x + x (- x) > 8$

Step 2.1.2.2

Rewrite using the commutative property of multiplication.

7 \cdot x - x \cdot x > 8

$7 \cdot x - x \cdot x > 8$

7 \cdot x - x \cdot x > 8

$7 \cdot x - x \cdot x > 8$

7 \cdot x - x \cdot x > 8

$7 \cdot x - x \cdot x > 8$

Step 2.2

Multiply

x

$x$ by

x

$x$ by adding the exponents.

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Step 2.2.1

Move

x

$x$ .

7 x - (x \cdot x) > 8

$7 x - (x \cdot x) > 8$

Step 2.2.2

Multiply

x

$x$ by

x

$x$ .

7 x - x^{2} > 8

$7 x - x^{2} > 8$

7 x - x^{2} > 8

$7 x - x^{2} > 8$

7 x - x^{2} > 8

$7 x - x^{2} > 8$

Step 3

Subtract

8

$8$ from both sides of the inequality.

7 x - x^{2} - 8 > 0

$7 x - x^{2} - 8 > 0$

Step 4

Convert the inequality to an equation.

7 x - x^{2} - 8 = 0

$7 x - x^{2} - 8 = 0$

Step 5

Use the quadratic formula to find the solutions.

\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 6

Substitute the values

a = - 1

$a = - 1$ ,

b = 7

$b = 7$ , and

c = - 8

$c = - 8$ into the quadratic formula and solve for

x

$x$ .

\frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (- 1 \cdot - 8)}}{2 \cdot - 1}

$\frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (- 1 \cdot - 8)}}{2 \cdot - 1}$

Step 7

Simplify.

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Step 7.1

Simplify the numerator.

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Step 7.1.1

Raise

7

$7$ to the power of

2

$2$ .

x = \frac{- 7 \pm \sqrt{49 - 4 \cdot - 1 \cdot - 8}}{2 \cdot - 1}

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot - 1 \cdot - 8}}{2 \cdot - 1}$

Step 7.1.2

Multiply

- 4 \cdot - 1 \cdot - 8

$- 4 \cdot - 1 \cdot - 8$ .

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Step 7.1.2.1

Multiply

- 4

$- 4$ by

- 1

$- 1$ .

x = \frac{- 7 \pm \sqrt{49 + 4 \cdot - 8}}{2 \cdot - 1}

$x = \frac{- 7 \pm \sqrt{49 + 4 \cdot - 8}}{2 \cdot - 1}$

Step 7.1.2.2

Multiply

4

$4$ by

- 8

$- 8$ .

x = \frac{- 7 \pm \sqrt{49 - 32}}{2 \cdot - 1}

$x = \frac{- 7 \pm \sqrt{49 - 32}}{2 \cdot - 1}$

x = \frac{- 7 \pm \sqrt{49 - 32}}{2 \cdot - 1}

$x = \frac{- 7 \pm \sqrt{49 - 32}}{2 \cdot - 1}$

Step 7.1.3

Subtract

32

$32$ from

49

$49$ .

x = \frac{- 7 \pm \sqrt{17}}{2 \cdot - 1}

$x = \frac{- 7 \pm \sqrt{17}}{2 \cdot - 1}$

x = \frac{- 7 \pm \sqrt{17}}{2 \cdot - 1}

$x = \frac{- 7 \pm \sqrt{17}}{2 \cdot - 1}$

Step 7.2

Multiply

2

$2$ by

- 1

$- 1$ .

x = \frac{- 7 \pm \sqrt{17}}{- 2}

$x = \frac{- 7 \pm \sqrt{17}}{- 2}$

Step 7.3

Simplify

\frac{- 7 \pm \sqrt{17}}{- 2}

$\frac{- 7 \pm \sqrt{17}}{- 2}$ .

x = \frac{7 \pm \sqrt{17}}{2}

$x = \frac{7 \pm \sqrt{17}}{2}$

x = \frac{7 \pm \sqrt{17}}{2}

$x = \frac{7 \pm \sqrt{17}}{2}$

Step 8

Consolidate the solutions.

x = \frac{7 + \sqrt{17}}{2}, \frac{7 - \sqrt{17}}{2}

$x = \frac{7 + \sqrt{17}}{2}, \frac{7 - \sqrt{17}}{2}$

Step 9

Use each root to create test intervals.

x < \frac{7 - \sqrt{17}}{2}

$x < \frac{7 - \sqrt{17}}{2}$

\frac{7 - \sqrt{17}}{2} < x < \frac{7 + \sqrt{17}}{2}

$\frac{7 - \sqrt{17}}{2} < x < \frac{7 + \sqrt{17}}{2}$

x > \frac{7 + \sqrt{17}}{2}

$x > \frac{7 + \sqrt{17}}{2}$

Step 10

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 10.1

Test a value on the interval

x < \frac{7 - \sqrt{17}}{2}

$x < \frac{7 - \sqrt{17}}{2}$ to see if it makes the inequality true.

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Step 10.1.1

Choose a value on the interval

x < \frac{7 - \sqrt{17}}{2}

$x < \frac{7 - \sqrt{17}}{2}$ and see if this value makes the original inequality true.

x = 0

$x = 0$

Step 10.1.2

Replace

x

$x$ with

0

$0$ in the original inequality.

8 < (0) (7 - (0))

$8 < (0) (7 - (0))$

Step 10.1.3

The left side

8

$8$ is not less than the right side

0

$0$ , which means that the given statement is false.

False

Step 10.2

Test a value on the interval

\frac{7 - \sqrt{17}}{2} < x < \frac{7 + \sqrt{17}}{2}

$\frac{7 - \sqrt{17}}{2} < x < \frac{7 + \sqrt{17}}{2}$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval

\frac{7 - \sqrt{17}}{2} < x < \frac{7 + \sqrt{17}}{2}

$\frac{7 - \sqrt{17}}{2} < x < \frac{7 + \sqrt{17}}{2}$ and see if this value makes the original inequality true.

x = 4

$x = 4$

Step 10.2.2

Replace

x

$x$ with

4

$4$ in the original inequality.

8 < (4) (7 - (4))

$8 < (4) (7 - (4))$

Step 10.2.3

The left side

8

$8$ is less than the right side

12

$12$ , which means that the given statement is always true.

True

Step 10.3

Test a value on the interval

x > \frac{7 + \sqrt{17}}{2}

$x > \frac{7 + \sqrt{17}}{2}$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval

x > \frac{7 + \sqrt{17}}{2}

$x > \frac{7 + \sqrt{17}}{2}$ and see if this value makes the original inequality true.

x = 8

$x = 8$

Step 10.3.2

Replace

x

$x$ with

8

$8$ in the original inequality.

8 < (8) (7 - (8))

$8 < (8) (7 - (8))$

Step 10.3.3

The left side

8

$8$ is not less than the right side

- 8

$- 8$ , which means that the given statement is false.

False

Step 10.4

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

x < \frac{7 - \sqrt{17}}{2}

$x < \frac{7 - \sqrt{17}}{2}$ False

\frac{7 - \sqrt{17}}{2} < x < \frac{7 + \sqrt{17}}{2}

$\frac{7 - \sqrt{17}}{2} < x < \frac{7 + \sqrt{17}}{2}$ True

x > \frac{7 + \sqrt{17}}{2}

$x > \frac{7 + \sqrt{17}}{2}$ False

x < \frac{7 - \sqrt{17}}{2}

$x < \frac{7 - \sqrt{17}}{2}$ False

\frac{7 - \sqrt{17}}{2} < x < \frac{7 + \sqrt{17}}{2}

$\frac{7 - \sqrt{17}}{2} < x < \frac{7 + \sqrt{17}}{2}$ True

x > \frac{7 + \sqrt{17}}{2}

$x > \frac{7 + \sqrt{17}}{2}$ False

Step 11

The solution consists of all of the true intervals.

\frac{7 - \sqrt{17}}{2} < x < \frac{7 + \sqrt{17}}{2}

$\frac{7 - \sqrt{17}}{2} < x < \frac{7 + \sqrt{17}}{2}$

Step 12

The result can be shown in multiple forms.

Inequality Form:

\frac{7 - \sqrt{17}}{2} < x < \frac{7 + \sqrt{17}}{2}

$\frac{7 - \sqrt{17}}{2} < x < \frac{7 + \sqrt{17}}{2}$

Interval Notation:

(\frac{7 - \sqrt{17}}{2}, \frac{7 + \sqrt{17}}{2})

$(\frac{7 - \sqrt{17}}{2}, \frac{7 + \sqrt{17}}{2})$

Step 13