Solve x^2-4x^2-4x=0

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

Solve for x

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

$$x=0$$

Steps Using Factoring

Factor out x.

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

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

$$x=0 x=-\frac{4}{3}$$

Steps Using the Quadratic Formula

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-4x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -4 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}}}{2\left(-3\right)}$$

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

$$x=\frac{-\left(-4\right)±4}{2\left(-3\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±4}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{4±4}{-6}$$

Now solve the equation x=\frac{4±4}{-6} when ± is plus. Add 4 to 4.

$$x=\frac{8}{-6}$$

Reduce the fraction \frac{8}{-6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{4}{3}$$

Now solve the equation x=\frac{4±4}{-6} when ± is minus. Subtract 4 from 4.

$$x=\frac{0}{-6}$$

Divide 0 by -6.

$$x=0$$

The equation is now solved.

$$x=-\frac{4}{3} x=0$$

Steps for Completing the Square

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-4x=0$$

Divide both sides by -3.

$$\frac{-3x^{2}-4x}{-3}=\frac{0}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{4}{-3}\right)x=\frac{0}{-3}$$

Divide -4 by -3.

$$x^{2}+\frac{4}{3}x=\frac{0}{-3}$$

Divide 0 by -3.

$$x^{2}+\frac{4}{3}x=0$$

Divide \frac{4}{3}, the coefficient of the x term, by 2 to get \frac{2}{3}. Then add the square of \frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{4}{3}x+\left(\frac{2}{3}\right)^{2}=\left(\frac{2}{3}\right)^{2}$$

Square \frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{4}{9}$$

Factor x^{2}+\frac{4}{3}x+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{2}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

$$x+\frac{2}{3}=\frac{2}{3} x+\frac{2}{3}=-\frac{2}{3}$$

Subtract \frac{2}{3} from both sides of the equation.

$$x=0 x=-\frac{4}{3}$$