Solve 8x^2=4x

$$8 { x }^{ 2 } = 4x$$

Solve for x

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

$$x=0$$

Steps Using Factoring

Subtract 4x from both sides.

$$8x^{2}-4x=0$$

Factor out x.

$$x\left(8x-4\right)=0$$

To find equation solutions, solve x=0 and 8x-4=0.

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

Steps Using the Quadratic Formula

Subtract 4x from both sides.

$$8x^{2}-4x=0$$

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

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}}}{2\times 8}$$

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

$$x=\frac{-\left(-4\right)±4}{2\times 8}$$

The opposite of -4 is 4.

$$x=\frac{4±4}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{4±4}{16}$$

Now solve the equation x=\frac{4±4}{16} when ± is plus. Add 4 to 4.

$$x=\frac{8}{16}$$

Reduce the fraction \frac{8}{16} to lowest terms by extracting and canceling out 8.

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

Now solve the equation x=\frac{4±4}{16} when ± is minus. Subtract 4 from 4.

$$x=\frac{0}{16}$$

Divide 0 by 16.

$$x=0$$

The equation is now solved.

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

Steps for Completing the Square

Subtract 4x from both sides.

$$8x^{2}-4x=0$$

Divide both sides by 8.

$$\frac{8x^{2}-4x}{8}=\frac{0}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\left(-\frac{4}{8}\right)x=\frac{0}{8}$$

Reduce the fraction \frac{-4}{8} to lowest terms by extracting and canceling out 4.

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

Divide 0 by 8.

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

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.

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