Solve x=8/8+x

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

Solve for x

$$x=2\sqrt{6}-4\approx 0.898979486$$

$$x=-2\sqrt{6}-4\approx -8.898979486$$

Steps Using the Quadratic Formula

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

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

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

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

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

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

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

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

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

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

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

Square 8.

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

Multiply -4 times -8.

$$x=\frac{-8±\sqrt{64+32}}{2}$$

Add 64 to 32.

$$x=\frac{-8±\sqrt{96}}{2}$$

Take the square root of 96.

$$x=\frac{-8±4\sqrt{6}}{2}$$

Now solve the equation x=\frac{-8±4\sqrt{6}}{2} when ± is plus. Add -8 to 4\sqrt{6}.

$$x=\frac{4\sqrt{6}-8}{2}$$

Divide -8+4\sqrt{6} by 2.

$$x=2\sqrt{6}-4$$

Now solve the equation x=\frac{-8±4\sqrt{6}}{2} when ± is minus. Subtract 4\sqrt{6} from -8.

$$x=\frac{-4\sqrt{6}-8}{2}$$

Divide -8-4\sqrt{6} by 2.

$$x=-2\sqrt{6}-4$$

The equation is now solved.

$$x=2\sqrt{6}-4 x=-2\sqrt{6}-4$$

Steps for Completing the Square

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

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

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

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

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

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

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

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

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

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

Add 8 to both sides. Anything plus zero gives itself.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

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.

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

Square 4.

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

Add 8 to 16.

$$x^{2}+8x+16=24$$

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}.

$$\left(x+4\right)^{2}=24$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{24}$$

Simplify.

$$x+4=2\sqrt{6} x+4=-2\sqrt{6}$$

Subtract 4 from both sides of the equation.

$$x=2\sqrt{6}-4 x=-2\sqrt{6}-4$$