Solve 1/3x=-x^2+2x

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

Solve for x

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

$$x=0$$

Steps Using Factoring

Add x^{2} to both sides.

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

Subtract 2x from both sides.

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

Combine \frac{1}{3}x and -2x to get -\frac{5}{3}x.

$$-\frac{5}{3}x+x^{2}=0$$

Factor out x.

$$x\left(-\frac{5}{3}+x\right)=0$$

To find equation solutions, solve x=0 and -\frac{5}{3}+x=0.

$$x=0 x=\frac{5}{3}$$

Steps Using the Quadratic Formula

Add x^{2} to both sides.

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

Subtract 2x from both sides.

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

Combine \frac{1}{3}x and -2x to get -\frac{5}{3}x.

$$-\frac{5}{3}x+x^{2}=0$$

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.

$$x^{2}-\frac{5}{3}x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -\frac{5}{3} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{5}{3}\right)±\sqrt{\left(-\frac{5}{3}\right)^{2}}}{2}$$

Take the square root of \left(-\frac{5}{3}\right)^{2}.

$$x=\frac{-\left(-\frac{5}{3}\right)±\frac{5}{3}}{2}$$

The opposite of -\frac{5}{3} is \frac{5}{3}.

$$x=\frac{\frac{5}{3}±\frac{5}{3}}{2}$$

Now solve the equation x=\frac{\frac{5}{3}±\frac{5}{3}}{2} when ± is plus. Add \frac{5}{3} to \frac{5}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{10}{3}}{2}$$

Divide \frac{10}{3} by 2.

$$x=\frac{5}{3}$$

Now solve the equation x=\frac{\frac{5}{3}±\frac{5}{3}}{2} when ± is minus. Subtract \frac{5}{3} from \frac{5}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide 0 by 2.

$$x=0$$

The equation is now solved.

$$x=\frac{5}{3} x=0$$

Steps for Completing the Square

Add x^{2} to both sides.

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

Subtract 2x from both sides.

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

Combine \frac{1}{3}x and -2x to get -\frac{5}{3}x.

$$-\frac{5}{3}x+x^{2}=0$$

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.

$$x^{2}-\frac{5}{3}x=0$$

Divide -\frac{5}{3}, the coefficient of the x term, by 2 to get -\frac{5}{6}. Then add the square of -\frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=\left(-\frac{5}{6}\right)^{2}$$

Square -\frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=\frac{25}{36}$$

Factor x^{2}-\frac{5}{3}x+\frac{25}{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}.

$$\left(x-\frac{5}{6}\right)^{2}=\frac{25}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Simplify.

$$x-\frac{5}{6}=\frac{5}{6} x-\frac{5}{6}=-\frac{5}{6}$$

Add \frac{5}{6} to both sides of the equation.

$$x=\frac{5}{3} x=0$$