Algebra Examples

$x (7 - x) > 8$

Step 1

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

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Step 1.1

Simplify $x (7 - x)$ .

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Step 1.1.1

Simplify by multiplying through.

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Step 1.1.1.1

Apply the distributive property.

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

Step 1.1.1.2

Reorder.

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Step 1.1.1.2.1

Move $7$ to the left of $x$ .

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

Step 1.1.1.2.2

Rewrite using the commutative property of multiplication.

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

Step 1.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 1.1.2.1

Move $x$ .

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

Step 1.1.2.2

Multiply $x$ by $x$ .

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

Step 1.2

Subtract $8$ from both sides of the inequality.

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

Step 1.3

Convert the inequality to an equation.

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

Step 1.4

Use the quadratic formula to find the solutions.

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

Step 1.5

Substitute the values $a = - 1$ , $b = 7$ , and $c = - 8$ into the quadratic formula and solve for $x$ .

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

Step 1.6

Simplify.

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Step 1.6.1

Simplify the numerator.

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Step 1.6.1.1

Raise $7$ to the power of $2$ .

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

Step 1.6.1.2

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

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Step 1.6.1.2.1

Multiply $- 4$ by $- 1$ .

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

Step 1.6.1.2.2

Multiply $4$ by $- 8$ .

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

Step 1.6.1.3

Subtract $32$ from $49$ .

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

Step 1.6.2

Multiply $2$ by $- 1$ .

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

Step 1.6.3

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

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

Step 1.7

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 1.7.1

Simplify the numerator.

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Step 1.7.1.1

Raise $7$ to the power of $2$ .

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

Step 1.7.1.2

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

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Step 1.7.1.2.1

Multiply $- 4$ by $- 1$ .

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

Step 1.7.1.2.2

Multiply $4$ by $- 8$ .

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

Step 1.7.1.3

Subtract $32$ from $49$ .

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

Step 1.7.2

Multiply $2$ by $- 1$ .

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

Step 1.7.3

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

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

Step 1.7.4

Change the $\pm$ to $+$ .

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

Step 1.8

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 1.8.1

Simplify the numerator.

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Step 1.8.1.1

Raise $7$ to the power of $2$ .

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

Step 1.8.1.2

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

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Step 1.8.1.2.1

Multiply $- 4$ by $- 1$ .

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

Step 1.8.1.2.2

Multiply $4$ by $- 8$ .

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

Step 1.8.1.3

Subtract $32$ from $49$ .

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

Step 1.8.2

Multiply $2$ by $- 1$ .

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

Step 1.8.3

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

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

Step 1.8.4

Change the $\pm$ to $-$ .

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

Step 1.9

Consolidate the solutions.

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

Step 1.10

Use each root to create test intervals.

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

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

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

Step 1.11

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 1.11.1

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

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Step 1.11.1.1

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

$x = 0$

Step 1.11.1.2

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

$(0) (7 - (0)) > 8$

Step 1.11.1.3

The left side $0$ is not greater than the right side $8$ , which means that the given statement is false.

False

Step 1.11.2

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

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Step 1.11.2.1

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

$x = 4$

Step 1.11.2.2

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

$(4) (7 - (4)) > 8$

Step 1.11.2.3

The left side $12$ is greater than the right side $8$ , which means that the given statement is always true.

True

Step 1.11.3

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

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Step 1.11.3.1

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

$x = 8$

Step 1.11.3.2

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

$(8) (7 - (8)) > 8$

Step 1.11.3.3

The left side $- 8$ is not greater than the right side $8$ , which means that the given statement is false.

False

Step 1.11.4

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

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

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

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

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

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

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

Step 1.12

The solution consists of all of the true intervals.

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

Step 2

Use the inequality $\frac{7 - \sqrt{17}}{2} < x < \frac{7 + \sqrt{17}}{2}$ to build the set notation.

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

Step 3