Solve 2x*1x=x

$$2x \times 1x=x$$

Solve for x

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

$$x=0$$

Steps Using Factoring

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

$$2x^{2}\times 1=x$$

Multiply 2 and 1 to get 2.

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

Subtract x from both sides.

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

Factor out x.

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

To find equation solutions, solve x=0 and 2x-1=0.

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

Steps Using the Quadratic Formula

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

$$2x^{2}\times 1=x$$

Multiply 2 and 1 to get 2.

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

Subtract x from both sides.

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

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

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

Take the square root of 1.

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

The opposite of -1 is 1.

$$x=\frac{1±1}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±1}{4}$$

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

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

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

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

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

$$x=\frac{0}{4}$$

Divide 0 by 4.

$$x=0$$

The equation is now solved.

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

Steps for Completing the Square

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

$$2x^{2}\times 1=x$$

Multiply 2 and 1 to get 2.

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

Subtract x from both sides.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 0 by 2.

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

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.

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

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

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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