Solve 8/8x=x+1

$$\frac { 8 } { 8 x } = x + 1$$

Solve for x

$$x=\frac{\sqrt{5}-1}{2}\approx 0.618033989$$

$$x=\frac{-\sqrt{5}-1}{2}\approx -1.618033989$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 8x.

$$8=8xx+8x$$

Multiply x and x to get x^{2}.

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

Swap sides so that all variable terms are on the left hand side.

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

Subtract 8 from both sides.

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

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

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

Square 8.

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

Multiply -4 times 8.

$$x=\frac{-8±\sqrt{64-32\left(-8\right)}}{2\times 8}$$

Multiply -32 times -8.

$$x=\frac{-8±\sqrt{64+256}}{2\times 8}$$

Add 64 to 256.

$$x=\frac{-8±\sqrt{320}}{2\times 8}$$

Take the square root of 320.

$$x=\frac{-8±8\sqrt{5}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-8±8\sqrt{5}}{16}$$

Now solve the equation x=\frac{-8±8\sqrt{5}}{16} when ± is plus. Add -8 to 8\sqrt{5}.

$$x=\frac{8\sqrt{5}-8}{16}$$

Divide -8+8\sqrt{5} by 16.

$$x=\frac{\sqrt{5}-1}{2}$$

Now solve the equation x=\frac{-8±8\sqrt{5}}{16} when ± is minus. Subtract 8\sqrt{5} from -8.

$$x=\frac{-8\sqrt{5}-8}{16}$$

Divide -8-8\sqrt{5} by 16.

$$x=\frac{-\sqrt{5}-1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{5}-1}{2} x=\frac{-\sqrt{5}-1}{2}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 8x.

$$8=8xx+8x$$

Multiply x and x to get x^{2}.

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

Swap sides so that all variable terms are on the left hand side.

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

Divide both sides by 8.

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

Dividing by 8 undoes the multiplication by 8.

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

Divide 8 by 8.

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

Divide 8 by 8.

$$x^{2}+x=1$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=1+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 1 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{5}{4}$$

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{5}}{2} x+\frac{1}{2}=-\frac{\sqrt{5}}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{5}-1}{2} x=\frac{-\sqrt{5}-1}{2}$$