Solve x^2=8x

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

Solve for x

$$x=8$$

$$x=0$$

Steps Using Factoring

Subtract 8x from both sides.

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

Factor out x.

$$x\left(x-8\right)=0$$

To find equation solutions, solve x=0 and x-8=0.

$$x=0 x=8$$

Steps Using the Quadratic Formula

Subtract 8x from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of \left(-8\right)^{2}.

$$x=\frac{-\left(-8\right)±8}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±8}{2}$$

Now solve the equation x=\frac{8±8}{2} when ± is plus. Add 8 to 8.

$$x=\frac{16}{2}$$

Divide 16 by 2.

$$x=8$$

Now solve the equation x=\frac{8±8}{2} when ± is minus. Subtract 8 from 8.

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

Divide 0 by 2.

$$x=0$$

The equation is now solved.

$$x=8 x=0$$

Steps for Completing the Square

Subtract 8x from both sides.

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

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.

$$x=8 x=0$$