Solve the Inequality for x -x^2-64<=-16x

$- x^{2} - 64 \leq - 16 x$

Step 1

Add $16 x$ to both sides of the inequality.

$- x^{2} - 64 + 16 x \leq 0$

Step 2

Convert the inequality to an equation.

$- x^{2} - 64 + 16 x = 0$

Step 3

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} - 64 + 16 x$ .

Move $- 64$ .

$- x^{2} + 16 x - 64 = 0$

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 16 x - 64 = 0$

Factor $- 1$ out of $16 x$ .

$- (x^{2}) - (- 16 x) - 64 = 0$

Rewrite $- 64$ as $- 1 (64)$ .

$- (x^{2}) - (- 16 x) - 1 \cdot 64 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 16 x)$ .

$- (x^{2} - 16 x) - 1 \cdot 64 = 0$

Factor $- 1$ out of $- (x^{2} - 16 x) - 1 (64)$ .

$- (x^{2} - 16 x + 64) = 0$

Factor using the perfect square rule.

Rewrite $64$ as $8^{2}$ .

$- (x^{2} - 16 x + 8^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$16 x = 2 \cdot x \cdot 8$

Rewrite the polynomial.

$- (x^{2} - 2 \cdot x \cdot 8 + 8^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 8$ .

$- {(x - 8)}^{2} = 0$

Step 4

Divide each term in $- {(x - 8)}^{2} = 0$ by $- 1$ and simplify.

Divide each term in $- {(x - 8)}^{2} = 0$ by $- 1$ .

$\frac{- {(x - 8)}^{2}}{- 1} = \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{{(x - 8)}^{2}}{1} = \frac{0}{- 1}$

Divide ${(x - 8)}^{2}$ by $1$ .

${(x - 8)}^{2} = \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

${(x - 8)}^{2} = 0$

Step 5

Set the $x - 8$ equal to $0$ .

$x - 8 = 0$

Step 6

Add $8$ to both sides of the equation.

$x = 8$

Step 7

Use each root to create test intervals.

$x < 8$

$x > 8$

Step 8

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 $x < 8$ to see if it makes the inequality true.

Choose a value on the interval $x < 8$ and see if this value makes the original inequality true.

$x = 0$

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

$- {(0)}^{2} - 64 \leq - 16 \cdot 0$

The left side $- 64$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 8$ to see if it makes the inequality true.

Choose a value on the interval $x > 8$ and see if this value makes the original inequality true.

$x = 10$

Replace $x$ with $10$ in the original inequality.

$- {(10)}^{2} - 64 \leq - 16 \cdot 10$

The left side $- 164$ is less than the right side $- 160$ , which means that the given statement is always true.

True

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

$x < 8$ True

$x > 8$ True

$x < 8$ True

$x > 8$ True

Step 9

The solution consists of all of the true intervals.

$x \leq 8$ or $x \geq 8$

Step 10

Combine the intervals.

All real numbers

Step 11

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$

Step 12

Solve the Inequality for x -x^2-64<=-16x

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