Solve {x=3x^2

$$\{ x = 3 x ^ { 2 }$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=0$$

Steps Using Factoring

Subtract 3x^{2} from both sides.

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

Factor out x.

$$x\left(1-3x\right)=0$$

To find equation solutions, solve x=0 and 1-3x=0.

$$x=0 x=\frac{1}{3}$$

Steps Using the Quadratic Formula

Subtract 3x^{2} from both sides.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 1 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}}}{2\left(-3\right)}$$

Take the square root of 1^{2}.

$$x=\frac{-1±1}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-1±1}{-6}$$

Now solve the equation x=\frac{-1±1}{-6} when ± is plus. Add -1 to 1.

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

Divide 0 by -6.

$$x=0$$

Now solve the equation x=\frac{-1±1}{-6} when ± is minus. Subtract 1 from -1.

$$x=-\frac{2}{-6}$$

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

$$x=\frac{1}{3}$$

The equation is now solved.

$$x=0 x=\frac{1}{3}$$

Steps for Completing the Square

Subtract 3x^{2} from both sides.

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

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.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{1}{-3}x=\frac{0}{-3}$$

Divide 1 by -3.

$$x^{2}-\frac{1}{3}x=\frac{0}{-3}$$

Divide 0 by -3.

$$x^{2}-\frac{1}{3}x=0$$

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{1}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{1}{6} x-\frac{1}{6}=-\frac{1}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$x=\frac{1}{3} x=0$$