Solve 1/x-2x-2=0

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

Solve for x

$$x=\frac{\sqrt{3}-1}{2}\approx 0.366025404$$

$$x=\frac{-\sqrt{3}-1}{2}\approx -1.366025404$$

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.

$$1-2xx+x\left(-2\right)=0$$

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

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

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.

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

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

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

Square -2.

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

Multiply -4 times -2.

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

Add 4 to 8.

$$x=\frac{-\left(-2\right)±\sqrt{12}}{2\left(-2\right)}$$

Take the square root of 12.

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

The opposite of -2 is 2.

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

Multiply 2 times -2.

$$x=\frac{2±2\sqrt{3}}{-4}$$

Now solve the equation x=\frac{2±2\sqrt{3}}{-4} when ± is plus. Add 2 to 2\sqrt{3}.

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

Divide 2+2\sqrt{3} by -4.

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

Now solve the equation x=\frac{2±2\sqrt{3}}{-4} when ± is minus. Subtract 2\sqrt{3} from 2.

$$x=\frac{2-2\sqrt{3}}{-4}$$

Divide 2-2\sqrt{3} by -4.

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

The equation is now solved.

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

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.

$$1-2xx+x\left(-2\right)=0$$

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

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

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

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

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.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -2 by -2.

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

Divide -1 by -2.

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

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.

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

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

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

Add \frac{1}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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.

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

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

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