Solve varphi=1/1+varphi

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

Solve for φ

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

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

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}.

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

Square 1.

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

Multiply -4 times -1.

$$\phi =\frac{-1±\sqrt{1+4}}{2}$$

Add 1 to 4.

$$\phi =\frac{-1±\sqrt{5}}{2}$$

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

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

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

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

The equation is now solved.

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

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 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}.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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