Solve 0=1/2x^2-2x

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

Solve for x

$$x=4$$

$$x=0$$

Steps Using Factoring

Swap sides so that all variable terms are on the left hand side.

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

Factor out x.

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

To find equation solutions, solve x=0 and \frac{x}{2}-2=0.

$$x=0 x=4$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

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

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

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

Take the square root of \left(-2\right)^{2}.

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

The opposite of -2 is 2.

$$x=\frac{2±2}{2\times \frac{1}{2}}$$

Multiply 2 times \frac{1}{2}.

$$x=\frac{2±2}{1}$$

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

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

Divide 4 by 1.

$$x=4$$

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

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

Divide 0 by 1.

$$x=0$$

The equation is now solved.

$$x=4 x=0$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

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

Multiply both sides by 2.

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

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

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

Divide -2 by \frac{1}{2} by multiplying -2 by the reciprocal of \frac{1}{2}.

$$x^{2}-4x=\frac{0}{\frac{1}{2}}$$

Divide 0 by \frac{1}{2} by multiplying 0 by the reciprocal of \frac{1}{2}.

$$x^{2}-4x=0$$

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

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

Square -2.

$$x^{2}-4x+4=4$$

Factor x^{2}-4x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=2 x-2=-2$$

Add 2 to both sides of the equation.

$$x=4 x=0$$