Solve 1/3x^2+3x+6geq0

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

Solve for x

$$x\in (-\infty,-6]\cup [-3,\infty)$$

Steps Using the Quadratic Formula

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.

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

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, 3 for b, and 6 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times \frac{1}{3}\times 6}}{\frac{1}{3}\times 2}$$

Do the calculations.

$$x=\frac{-3±1}{\frac{2}{3}}$$

Solve the equation x=\frac{-3±1}{\frac{2}{3}} when ± is plus and when ± is minus.

$$x=-3 x=-6$$

Rewrite the inequality by using the obtained solutions.

$$\frac{1}{3}\left(x+3\right)\left(x+6\right)\geq 0$$

For the product to be ≥0, x+3 and x+6 have to be both ≤0 or both ≥0. Consider the case when x+3 and x+6 are both ≤0.

$$x+3\leq 0 x+6\leq 0$$

The solution satisfying both inequalities is x\leq -6.

$$x\leq -6$$

Consider the case when x+3 and x+6 are both ≥0.

$$x+6\geq 0 x+3\geq 0$$

The solution satisfying both inequalities is x\geq -3.

$$x\geq -3$$

The final solution is the union of the obtained solutions.

$$x\leq -6\text{; }x\geq -3$$