Solve 1/x+x=1

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

Solve for x (complex solution)

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

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

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 x.

$$1+xx=x$$

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

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

Subtract x from both sides.

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

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}.

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

Add 1 to -4.

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

Take the square root of -3.

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

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{3}i}{2}$$

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

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

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

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

The equation is now solved.

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

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 x.

$$1+xx=x$$

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

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

Subtract x from both sides.

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

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

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

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}.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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