Solve 2x-1=1/x

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

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=1$$

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.

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

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

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

Subtract 1 from both sides.

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

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.

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

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

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

Multiply -4 times 2.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\left(-1\right)}}{2\times 2}$$

Multiply -8 times -1.

$$x=\frac{-\left(-1\right)±\sqrt{1+8}}{2\times 2}$$

Add 1 to 8.

$$x=\frac{-\left(-1\right)±\sqrt{9}}{2\times 2}$$

Take the square root of 9.

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

The opposite of -1 is 1.

$$x=\frac{1±3}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±3}{4}$$

Now solve the equation x=\frac{1±3}{4} when ± is plus. Add 1 to 3.

$$x=\frac{4}{4}$$

Divide 4 by 4.

$$x=1$$

Now solve the equation x=\frac{1±3}{4} when ± is minus. Subtract 3 from 1.

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

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

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.

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

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

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

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.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1}{2}+\frac{1}{16}$$

Add \frac{1}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{9}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{3}{4} x-\frac{1}{4}=-\frac{3}{4}$$

Add \frac{1}{4} to both sides of the equation.

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