Solve the Inequality for x 1/8x^2+x<=-2

$\frac{1}{8} x^{2} + x \leq - 2$

Step 1

Combine $\frac{1}{8}$ and $x^{2}$ .

$\frac{x^{2}}{8} + x \leq - 2$

Step 2

Multiply each term in $\frac{x^{2}}{8} + x \leq - 2$ by $8$ to eliminate the fractions.

Multiply each term in $\frac{x^{2}}{8} + x \leq - 2$ by $8$ .

$\frac{x^{2}}{8} \cdot 8 + x \cdot 8 \leq - 2 \cdot 8$

Simplify the left side.

Simplify each term.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{x^{2}}{8} \cdot 8 + x \cdot 8 \leq - 2 \cdot 8$

Rewrite the expression.

$x^{2} + x \cdot 8 \leq - 2 \cdot 8$

Move $8$ to the left of $x$ .

$x^{2} + 8 x \leq - 2 \cdot 8$

Simplify the right side.

Multiply $- 2$ by $8$ .

$x^{2} + 8 x \leq - 16$

Step 3

Convert the inequality to an equation.

$x^{2} + 8 x = - 16$

Step 4

Add $16$ to both sides of the equation.

$x^{2} + 8 x + 16 = 0$

Step 5

Factor using the perfect square rule.

Rewrite $16$ as $4^{2}$ .

$x^{2} + 8 x + 4^{2} = 0$

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

$8 x = 2 \cdot x \cdot 4$

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 4 + 4^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 4$ .

${(x + 4)}^{2} = 0$

Step 6

Set the $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 7

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 8