Solve 1+1/1+1{1+frac{1/x}}=x

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

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

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

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

Since \frac{x}{x} and \frac{1}{x} have the same denominator, add them by adding their numerators.

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

Variable x cannot be equal to 0 since division by zero is not defined. Divide 1 by \frac{x+1}{x} by multiplying 1 by the reciprocal of \frac{x+1}{x}.

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

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

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

Since \frac{x+1}{x+1} and \frac{x}{x+1} have the same denominator, add them by adding their numerators.

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

Combine like terms in x+1+x.

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

Variable x cannot be equal to -1 since division by zero is not defined. Divide 1 by \frac{2x+1}{x+1} by multiplying 1 by the reciprocal of \frac{2x+1}{x+1}.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2x+1}{2x+1}.

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

Since \frac{2x+1}{2x+1} and \frac{x+1}{2x+1} have the same denominator, add them by adding their numerators.

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

Combine like terms in 2x+1+x+1.

$$\frac{3x+2}{2x+1}=x$$

Subtract x from both sides.

$$\frac{3x+2}{2x+1}-x=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{2x+1}{2x+1}.

$$\frac{3x+2}{2x+1}-\frac{x\left(2x+1\right)}{2x+1}=0$$

Since \frac{3x+2}{2x+1} and \frac{x\left(2x+1\right)}{2x+1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3x+2-x\left(2x+1\right)}{2x+1}=0$$

Do the multiplications in 3x+2-x\left(2x+1\right).

$$\frac{3x+2-2x^{2}-x}{2x+1}=0$$

Combine like terms in 3x+2-2x^{2}-x.

$$\frac{2x+2-2x^{2}}{2x+1}=0$$

Variable x cannot be equal to -\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by 2x+1.

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

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.

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

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

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

Square 2.

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

Multiply -4 times -2.

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

Multiply 8 times 2.

$$x=\frac{-2±\sqrt{4+16}}{2\left(-2\right)}$$

Add 4 to 16.

$$x=\frac{-2±\sqrt{20}}{2\left(-2\right)}$$

Take the square root of 20.

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

Multiply 2 times -2.

$$x=\frac{-2±2\sqrt{5}}{-4}$$

Now solve the equation x=\frac{-2±2\sqrt{5}}{-4} when ± is plus. Add -2 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-2}{-4}$$

Divide -2+2\sqrt{5} by -4.

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

Now solve the equation x=\frac{-2±2\sqrt{5}}{-4} when ± is minus. Subtract 2\sqrt{5} from -2.

$$x=\frac{-2\sqrt{5}-2}{-4}$$

Divide -2-2\sqrt{5} by -4.

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

The equation is now solved.

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

Steps for Completing the Square

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

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

Since \frac{x}{x} and \frac{1}{x} have the same denominator, add them by adding their numerators.

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

Variable x cannot be equal to 0 since division by zero is not defined. Divide 1 by \frac{x+1}{x} by multiplying 1 by the reciprocal of \frac{x+1}{x}.

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

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

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

Since \frac{x+1}{x+1} and \frac{x}{x+1} have the same denominator, add them by adding their numerators.

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

Combine like terms in x+1+x.

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

Variable x cannot be equal to -1 since division by zero is not defined. Divide 1 by \frac{2x+1}{x+1} by multiplying 1 by the reciprocal of \frac{2x+1}{x+1}.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2x+1}{2x+1}.

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

Since \frac{2x+1}{2x+1} and \frac{x+1}{2x+1} have the same denominator, add them by adding their numerators.

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

Combine like terms in 2x+1+x+1.

$$\frac{3x+2}{2x+1}=x$$

Subtract x from both sides.

$$\frac{3x+2}{2x+1}-x=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{2x+1}{2x+1}.

$$\frac{3x+2}{2x+1}-\frac{x\left(2x+1\right)}{2x+1}=0$$

Since \frac{3x+2}{2x+1} and \frac{x\left(2x+1\right)}{2x+1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3x+2-x\left(2x+1\right)}{2x+1}=0$$

Do the multiplications in 3x+2-x\left(2x+1\right).

$$\frac{3x+2-2x^{2}-x}{2x+1}=0$$

Combine like terms in 3x+2-2x^{2}-x.

$$\frac{2x+2-2x^{2}}{2x+1}=0$$

Variable x cannot be equal to -\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by 2x+1.

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

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 2 by -2.

$$x^{2}-x=-\frac{2}{-2}$$

Divide -2 by -2.

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

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.

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