Solve 1/3x^2+4x=2x

$$\frac { 1 } { 3 } x ^ { 2 } + 4 x = 2 x$$

Solve for x

$$x=-6$$

$$x=0$$

Steps Using Factoring

Subtract 2x from both sides.

$$\frac{1}{3}x^{2}+4x-2x=0$$

Combine 4x and -2x to get 2x.

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

Factor out x.

$$x\left(\frac{1}{3}x+2\right)=0$$

To find equation solutions, solve x=0 and \frac{x}{3}+2=0.

$$x=0 x=-6$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

$$\frac{1}{3}x^{2}+4x-2x=0$$

Combine 4x and -2x to get 2x.

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

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

$$x=\frac{-2±\sqrt{2^{2}}}{2\times \frac{1}{3}}$$

Take the square root of 2^{2}.

$$x=\frac{-2±2}{2\times \frac{1}{3}}$$

Multiply 2 times \frac{1}{3}.

$$x=\frac{-2±2}{\frac{2}{3}}$$

Now solve the equation x=\frac{-2±2}{\frac{2}{3}} when ± is plus. Add -2 to 2.

$$x=\frac{0}{\frac{2}{3}}$$

Divide 0 by \frac{2}{3} by multiplying 0 by the reciprocal of \frac{2}{3}.

$$x=0$$

Now solve the equation x=\frac{-2±2}{\frac{2}{3}} when ± is minus. Subtract 2 from -2.

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

Divide -4 by \frac{2}{3} by multiplying -4 by the reciprocal of \frac{2}{3}.

$$x=-6$$

The equation is now solved.

$$x=0 x=-6$$

Steps for Completing the Square

Subtract 2x from both sides.

$$\frac{1}{3}x^{2}+4x-2x=0$$

Combine 4x and -2x to get 2x.

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

Multiply both sides by 3.

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

Dividing by \frac{1}{3} undoes the multiplication by \frac{1}{3}.

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

Divide 2 by \frac{1}{3} by multiplying 2 by the reciprocal of \frac{1}{3}.

$$x^{2}+6x=\frac{0}{\frac{1}{3}}$$

Divide 0 by \frac{1}{3} by multiplying 0 by the reciprocal of \frac{1}{3}.

$$x^{2}+6x=0$$

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

$$x^{2}+6x+3^{2}=3^{2}$$

Square 3.

$$x^{2}+6x+9=9$$

Factor x^{2}+6x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+3=3 x+3=-3$$

Subtract 3 from both sides of the equation.

$$x=0 x=-6$$