Solve x=2/x+8

$$x= \frac{ 2 }{ x+8 }$$

Solve for x

$$x=3\sqrt{2}-4\approx 0.242640687$$

$$x=-3\sqrt{2}-4\approx -8.242640687$$

Steps Using the Quadratic Formula

Subtract \frac{2}{x+8} from both sides.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x+8}{x+8}.

$$\frac{x\left(x+8\right)}{x+8}-\frac{2}{x+8}=0$$

Since \frac{x\left(x+8\right)}{x+8} and \frac{2}{x+8} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x+8\right)-2}{x+8}=0$$

Do the multiplications in x\left(x+8\right)-2.

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

Variable x cannot be equal to -8 since division by zero is not defined. Multiply both sides of the equation by x+8.

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

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

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

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-2\right)}}{2}$$

Multiply -4 times -2.

$$x=\frac{-8±\sqrt{64+8}}{2}$$

Add 64 to 8.

$$x=\frac{-8±\sqrt{72}}{2}$$

Take the square root of 72.

$$x=\frac{-8±6\sqrt{2}}{2}$$

Now solve the equation x=\frac{-8±6\sqrt{2}}{2} when ± is plus. Add -8 to 6\sqrt{2}.

$$x=\frac{6\sqrt{2}-8}{2}$$

Divide -8+6\sqrt{2} by 2.

$$x=3\sqrt{2}-4$$

Now solve the equation x=\frac{-8±6\sqrt{2}}{2} when ± is minus. Subtract 6\sqrt{2} from -8.

$$x=\frac{-6\sqrt{2}-8}{2}$$

Divide -8-6\sqrt{2} by 2.

$$x=-3\sqrt{2}-4$$

The equation is now solved.

$$x=3\sqrt{2}-4 x=-3\sqrt{2}-4$$

Steps for Completing the Square

Subtract \frac{2}{x+8} from both sides.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x+8}{x+8}.

$$\frac{x\left(x+8\right)}{x+8}-\frac{2}{x+8}=0$$

Since \frac{x\left(x+8\right)}{x+8} and \frac{2}{x+8} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(x+8\right)-2}{x+8}=0$$

Do the multiplications in x\left(x+8\right)-2.

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

Variable x cannot be equal to -8 since division by zero is not defined. Multiply both sides of the equation by x+8.

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

Add 2 to both sides. Anything plus zero gives itself.

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

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square 4.

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

Add 2 to 16.

$$x^{2}+8x+16=18$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+4\right)^{2}=18$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{18}$$

Simplify.

$$x+4=3\sqrt{2} x+4=-3\sqrt{2}$$

Subtract 4 from both sides of the equation.

$$x=3\sqrt{2}-4 x=-3\sqrt{2}-4$$