Solve 0=4x-1/2x^2

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

Solve for x

$$x=8$$

$$x=0$$

Steps Using Factoring

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

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

Factor out x.

$$x\left(4-\frac{1}{2}x\right)=0$$

To find equation solutions, solve x=0 and 4-\frac{x}{2}=0.

$$x=0 x=8$$

Steps Using the Quadratic Formula

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

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

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.

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

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

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

Take the square root of 4^{2}.

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

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

$$x=\frac{-4±4}{-1}$$

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

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

Divide 0 by -1.

$$x=0$$

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

$$x=-\frac{8}{-1}$$

Divide -8 by -1.

$$x=8$$

The equation is now solved.

$$x=0 x=8$$

Steps for Completing the Square

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

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

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.

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

Multiply both sides by -2.

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

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

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

Divide 4 by -\frac{1}{2} by multiplying 4 by the reciprocal of -\frac{1}{2}.

$$x^{2}-8x=\frac{0}{-\frac{1}{2}}$$

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

$$x^{2}-8x=0$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=16$$

Factor x^{2}-8x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-4=4 x-4=-4$$

Add 4 to both sides of the equation.

$$x=8 x=0$$