Solve -1/8x^2+1/2x+4=-x-4

$$- \frac{ 1 }{ 8 } { x }^{ 2 } + \frac{ 1 }{ 2 } x+4 = -x-4$$

Solve for x

$$x=-4$$

$$x=16$$

Steps Using the Quadratic Formula

Add x to both sides.

$$-\frac{1}{8}x^{2}+\frac{1}{2}x+4+x=-4$$

Combine \frac{1}{2}x and x to get \frac{3}{2}x.

$$-\frac{1}{8}x^{2}+\frac{3}{2}x+4=-4$$

Add 4 to both sides.

$$-\frac{1}{8}x^{2}+\frac{3}{2}x+4+4=0$$

Add 4 and 4 to get 8.

$$-\frac{1}{8}x^{2}+\frac{3}{2}x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{8} for a, \frac{3}{2} for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{3}{2}±\sqrt{\left(\frac{3}{2}\right)^{2}-4\left(-\frac{1}{8}\right)\times 8}}{2\left(-\frac{1}{8}\right)}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}-4\left(-\frac{1}{8}\right)\times 8}}{2\left(-\frac{1}{8}\right)}$$

Multiply -4 times -\frac{1}{8}.

$$x=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}+\frac{1}{2}\times 8}}{2\left(-\frac{1}{8}\right)}$$

Multiply \frac{1}{2} times 8.

$$x=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}+4}}{2\left(-\frac{1}{8}\right)}$$

Add \frac{9}{4} to 4.

$$x=\frac{-\frac{3}{2}±\sqrt{\frac{25}{4}}}{2\left(-\frac{1}{8}\right)}$$

Take the square root of \frac{25}{4}.

$$x=\frac{-\frac{3}{2}±\frac{5}{2}}{2\left(-\frac{1}{8}\right)}$$

Multiply 2 times -\frac{1}{8}.

$$x=\frac{-\frac{3}{2}±\frac{5}{2}}{-\frac{1}{4}}$$

Now solve the equation x=\frac{-\frac{3}{2}±\frac{5}{2}}{-\frac{1}{4}} when ± is plus. Add -\frac{3}{2} to \frac{5}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{1}{-\frac{1}{4}}$$

Divide 1 by -\frac{1}{4} by multiplying 1 by the reciprocal of -\frac{1}{4}.

$$x=-4$$

Now solve the equation x=\frac{-\frac{3}{2}±\frac{5}{2}}{-\frac{1}{4}} when ± is minus. Subtract \frac{5}{2} from -\frac{3}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{4}{-\frac{1}{4}}$$

Divide -4 by -\frac{1}{4} by multiplying -4 by the reciprocal of -\frac{1}{4}.

$$x=16$$

The equation is now solved.

$$x=-4 x=16$$

Steps for Completing the Square

Add x to both sides.

$$-\frac{1}{8}x^{2}+\frac{1}{2}x+4+x=-4$$

Combine \frac{1}{2}x and x to get \frac{3}{2}x.

$$-\frac{1}{8}x^{2}+\frac{3}{2}x+4=-4$$

Subtract 4 from both sides.

$$-\frac{1}{8}x^{2}+\frac{3}{2}x=-4-4$$

Subtract 4 from -4 to get -8.

$$-\frac{1}{8}x^{2}+\frac{3}{2}x=-8$$

Multiply both sides by -8.

$$\frac{-\frac{1}{8}x^{2}+\frac{3}{2}x}{-\frac{1}{8}}=-\frac{8}{-\frac{1}{8}}$$

Dividing by -\frac{1}{8} undoes the multiplication by -\frac{1}{8}.

$$x^{2}+\frac{\frac{3}{2}}{-\frac{1}{8}}x=-\frac{8}{-\frac{1}{8}}$$

Divide \frac{3}{2} by -\frac{1}{8} by multiplying \frac{3}{2} by the reciprocal of -\frac{1}{8}.

$$x^{2}-12x=-\frac{8}{-\frac{1}{8}}$$

Divide -8 by -\frac{1}{8} by multiplying -8 by the reciprocal of -\frac{1}{8}.

$$x^{2}-12x=64$$

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-12x+\left(-6\right)^{2}=64+\left(-6\right)^{2}$$

Square -6.

$$x^{2}-12x+36=64+36$$

Add 64 to 36.

$$x^{2}-12x+36=100$$

Factor x^{2}-12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-6\right)^{2}=100$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{100}$$

Simplify.

$$x-6=10 x-6=-10$$

Add 6 to both sides of the equation.

$$x=16 x=-4$$