Solve x-1=x^-1

$$x - 1 = x ^ { - 1 }$$

Solve for x

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

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

Steps Using the Quadratic Formula

Subtract x^{-1} from both sides.

$$x-1-x^{-1}=0$$

Reorder the terms.

$$x-1-\frac{1}{x}=0$$

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

$$xx+x\left(-1\right)-1=0$$

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

$$x^{2}+x\left(-1\right)-1=0$$

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.

$$x^{2}-x-1=0$$

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

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

Multiply -4 times -1.

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

Add 1 to 4.

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

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{5}}{2}$$

Now solve the equation x=\frac{1±\sqrt{5}}{2} when ± is plus. Add 1 to \sqrt{5}.

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

Now solve the equation x=\frac{1±\sqrt{5}}{2} when ± is minus. Subtract \sqrt{5} from 1.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract x^{-1} from both sides.

$$x-1-x^{-1}=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$x-x^{-1}=1$$

Reorder the terms.

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

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

$$xx-1=x$$

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

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

Subtract x from both sides.

$$x^{2}-1-x=0$$

Add 1 to both sides. Anything plus zero gives itself.

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

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.

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

Add \frac{1}{2} to both sides of the equation.

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