Solve 16x^2-4x=0

$$16 x ^ { 2 } - 4 x = 0$$

Solve for x

$$x=\frac{1}{4}=0.25$$

$$x=0$$

Steps Using Factoring

Factor out x.

$$x\left(16x-4\right)=0$$

To find equation solutions, solve x=0 and 16x-4=0.

$$x=0 x=\frac{1}{4}$$

Steps Using the Quadratic Formula

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.

$$16x^{2}-4x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -4 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

$$x=\frac{-\left(-4\right)±4}{2\times 16}$$

The opposite of -4 is 4.

$$x=\frac{4±4}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{4±4}{32}$$

Now solve the equation x=\frac{4±4}{32} when ± is plus. Add 4 to 4.

$$x=\frac{8}{32}$$

Reduce the fraction \frac{8}{32} to lowest terms by extracting and canceling out 8.

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

Now solve the equation x=\frac{4±4}{32} when ± is minus. Subtract 4 from 4.

$$x=\frac{0}{32}$$

Divide 0 by 32.

$$x=0$$

The equation is now solved.

$$x=\frac{1}{4} x=0$$

Steps for Completing the Square

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.

$$16x^{2}-4x=0$$

Divide both sides by 16.

$$\frac{16x^{2}-4x}{16}=\frac{0}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\left(-\frac{4}{16}\right)x=\frac{0}{16}$$

Reduce the fraction \frac{-4}{16} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{1}{4}x=\frac{0}{16}$$

Divide 0 by 16.

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

Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{4}x+\left(-\frac{1}{8}\right)^{2}=\left(-\frac{1}{8}\right)^{2}$$

Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{1}{64}$$

Factor x^{2}-\frac{1}{4}x+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{8}\right)^{2}=\frac{1}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Simplify.

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

Add \frac{1}{8} to both sides of the equation.

$$x=\frac{1}{4} x=0$$