Solve -1/3(x+2)(x-1/3)>0

$$- \frac { 1 } { 3 } ( x + 2 ) ( x - \frac { 1 } { 3 } ) > 0$$

Solve for x

$$x\in \left(-2,\frac{1}{3}\right)$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -\frac{1}{3} by x+2.

$$\left(-\frac{1}{3}x-\frac{2}{3}\right)\left(x-\frac{1}{3}\right)>0$$

Use the distributive property to multiply -\frac{1}{3}x-\frac{2}{3} by x-\frac{1}{3} and combine like terms.

$$-\frac{1}{3}x^{2}-\frac{5}{9}x+\frac{2}{9}>0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -\frac{1}{3}x^{2}-\frac{5}{9}x+\frac{2}{9} positive. Since -1 is negative, the inequality direction is changed.

$$\frac{1}{3}x^{2}+\frac{5}{9}x-\frac{2}{9}<0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$\frac{1}{3}x^{2}+\frac{5}{9}x-\frac{2}{9}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute \frac{1}{3} for a, \frac{5}{9} for b, and -\frac{2}{9} for c in the quadratic formula.

$$x=\frac{-\frac{5}{9}±\sqrt{\left(\frac{5}{9}\right)^{2}-4\times \frac{1}{3}\left(-\frac{2}{9}\right)}}{\frac{1}{3}\times 2}$$

Do the calculations.

$$x=\frac{-\frac{5}{9}±\frac{7}{9}}{\frac{2}{3}}$$

Solve the equation x=\frac{-\frac{5}{9}±\frac{7}{9}}{\frac{2}{3}} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

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

For the product to be negative, x-\frac{1}{3} and x+2 have to be of the opposite signs. Consider the case when x-\frac{1}{3} is positive and x+2 is negative.

$$x-\frac{1}{3}>0 x+2<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+2 is positive and x-\frac{1}{3} is negative.

$$x+2>0 x-\frac{1}{3}<0$$

The solution satisfying both inequalities is x\in \left(-2,\frac{1}{3}\right).

$$x\in \left(-2,\frac{1}{3}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-2,\frac{1}{3}\right)$$