Solve varphi=1/1-varphi

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

Solve for φ

$$\phi =\frac{-\sqrt{3}i+1}{2}\approx 0.5-0.866025404i$$

$$\phi =\frac{1+\sqrt{3}i}{2}\approx 0.5+0.866025404i$$

Steps Using the Quadratic Formula

Subtract \frac{1}{1-\phi } from both sides.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply \phi times \frac{1-\phi }{1-\phi }.

$$\frac{\phi \left(1-\phi \right)}{1-\phi }-\frac{1}{1-\phi }=0$$

Since \frac{\phi \left(1-\phi \right)}{1-\phi } and \frac{1}{1-\phi } have the same denominator, subtract them by subtracting their numerators.

$$\frac{\phi \left(1-\phi \right)-1}{1-\phi }=0$$

Do the multiplications in \phi \left(1-\phi \right)-1.

$$\frac{\phi -\phi ^{2}-1}{1-\phi }=0$$

Variable \phi cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by -\phi +1.

$$\phi -\phi ^{2}-1=0$$

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.

$$-\phi ^{2}+\phi -1=0$$

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}.

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

Square 1.

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

Multiply -4 times -1.

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

Multiply 4 times -1.

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

Add 1 to -4.

$$\phi =\frac{-1±\sqrt{-3}}{2\left(-1\right)}$$

Take the square root of -3.

$$\phi =\frac{-1±\sqrt{3}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$\phi =\frac{-1±\sqrt{3}i}{-2}$$

Now solve the equation \phi =\frac{-1±\sqrt{3}i}{-2} when ± is plus. Add -1 to i\sqrt{3}.

$$\phi =\frac{-1+\sqrt{3}i}{-2}$$

Divide -1+i\sqrt{3} by -2.

$$\phi =\frac{-\sqrt{3}i+1}{2}$$

Now solve the equation \phi =\frac{-1±\sqrt{3}i}{-2} when ± is minus. Subtract i\sqrt{3} from -1.

$$\phi =\frac{-\sqrt{3}i-1}{-2}$$

Divide -1-i\sqrt{3} by -2.

$$\phi =\frac{1+\sqrt{3}i}{2}$$

The equation is now solved.

$$\phi =\frac{-\sqrt{3}i+1}{2} \phi =\frac{1+\sqrt{3}i}{2}$$

Steps for Completing the Square

Subtract \frac{1}{1-\phi } from both sides.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply \phi times \frac{1-\phi }{1-\phi }.

$$\frac{\phi \left(1-\phi \right)}{1-\phi }-\frac{1}{1-\phi }=0$$

Since \frac{\phi \left(1-\phi \right)}{1-\phi } and \frac{1}{1-\phi } have the same denominator, subtract them by subtracting their numerators.

$$\frac{\phi \left(1-\phi \right)-1}{1-\phi }=0$$

Do the multiplications in \phi \left(1-\phi \right)-1.

$$\frac{\phi -\phi ^{2}-1}{1-\phi }=0$$

Variable \phi cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by -\phi +1.

$$\phi -\phi ^{2}-1=0$$

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

$$\phi -\phi ^{2}=1$$

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.

$$-\phi ^{2}+\phi =1$$

Divide both sides by -1.

$$\frac{-\phi ^{2}+\phi }{-1}=\frac{1}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$\phi ^{2}+\frac{1}{-1}\phi =\frac{1}{-1}$$

Divide 1 by -1.

$$\phi ^{2}-\phi =\frac{1}{-1}$$

Divide 1 by -1.

$$\phi ^{2}-\phi =-1$$

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.

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

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

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

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

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

Factor \phi ^{2}-\phi +\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}.

$$\left(\phi -\frac{1}{2}\right)^{2}=-\frac{3}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(\phi -\frac{1}{2}\right)^{2}}=\sqrt{-\frac{3}{4}}$$

Simplify.

$$\phi -\frac{1}{2}=\frac{\sqrt{3}i}{2} \phi -\frac{1}{2}=-\frac{\sqrt{3}i}{2}$$

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

$$\phi =\frac{1+\sqrt{3}i}{2} \phi =\frac{-\sqrt{3}i+1}{2}$$