Solve for x 8<x(7-x)

$8 < x (7 - x)$

Step 1

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

$x (7 - x) > 8$

Step 2

Simplify $x (7 - x)$ .

Simplify by multiplying through.

Apply the distributive property.

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

Reorder.

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

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

Rewrite using the commutative property of multiplication.

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

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

Move $x$ .

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

Multiply $x$ by $x$ .

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

Step 3

Subtract $8$ from both sides of the inequality.

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

Step 4

Convert the inequality to an equation.

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

Step 5

Use the quadratic formula to find the solutions.

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

Step 6

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 7

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $- 1$ .

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

Multiply $4$ by $- 8$ .

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

Subtract $32$ from $49$ .

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

Multiply $2$ by $- 1$ .

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

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

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

Step 8

Consolidate the solutions.

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

Step 9

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 10

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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

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

$x = 0$

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

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

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

False

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.

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.

$x = 4$

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

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

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

True

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

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

$x = 8$

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

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

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

False

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 11

The solution consists of all of the true intervals.

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

Step 12

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

Step 13

Solve for x 8<x(7-x)

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