Solve 1000x^2-500x=0

$$1000 x ^ { 2 } - 500 x = 0$$

Solve for x

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

$$x=0$$

Steps Using Factoring

Factor out x.

$$x\left(1000x-500\right)=0$$

To find equation solutions, solve x=0 and 1000x-500=0.

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

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.

$$1000x^{2}-500x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1000 for a, -500 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-500\right)±\sqrt{\left(-500\right)^{2}}}{2\times 1000}$$

Take the square root of \left(-500\right)^{2}.

$$x=\frac{-\left(-500\right)±500}{2\times 1000}$$

The opposite of -500 is 500.

$$x=\frac{500±500}{2\times 1000}$$

Multiply 2 times 1000.

$$x=\frac{500±500}{2000}$$

Now solve the equation x=\frac{500±500}{2000} when ± is plus. Add 500 to 500.

$$x=\frac{1000}{2000}$$

Reduce the fraction \frac{1000}{2000} to lowest terms by extracting and canceling out 1000.

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

Now solve the equation x=\frac{500±500}{2000} when ± is minus. Subtract 500 from 500.

$$x=\frac{0}{2000}$$

Divide 0 by 2000.

$$x=0$$

The equation is now solved.

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

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.

$$1000x^{2}-500x=0$$

Divide both sides by 1000.

$$\frac{1000x^{2}-500x}{1000}=\frac{0}{1000}$$

Dividing by 1000 undoes the multiplication by 1000.

$$x^{2}+\left(-\frac{500}{1000}\right)x=\frac{0}{1000}$$

Reduce the fraction \frac{-500}{1000} to lowest terms by extracting and canceling out 500.

$$x^{2}-\frac{1}{2}x=\frac{0}{1000}$$

Divide 0 by 1000.

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

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.

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