Solve x^2=4x

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

Solve for x

$$x=4$$

$$x=0$$

Steps Using Factoring

Subtract 4x from both sides.

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

Factor out x.

$$x\left(x-4\right)=0$$

To find equation solutions, solve x=0 and x-4=0.

$$x=0 x=4$$

Steps Using the Quadratic Formula

Subtract 4x from both sides.

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

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

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

Take the square root of \left(-4\right)^{2}.

$$x=\frac{-\left(-4\right)±4}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±4}{2}$$

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

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

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

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

Divide 0 by 2.

$$x=0$$

The equation is now solved.

$$x=4 x=0$$

Steps for Completing the Square

Subtract 4x from both sides.

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

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.

$$x=4 x=0$$