Solve u^2+6u+6leq0

$$u ^ { 2 } + 6 u + 6 \leq 0$$

Solve for u

$$u\in \begin{bmatrix}-\sqrt{3}-3,\sqrt{3}-3\end{bmatrix}$$

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.

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

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 1 for a, 6 for b, and 6 for c in the quadratic formula.

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

Do the calculations.

$$u=\frac{-6±2\sqrt{3}}{2}$$

Solve the equation u=\frac{-6±2\sqrt{3}}{2} when ± is plus and when ± is minus.

$$u=\sqrt{3}-3 u=-\sqrt{3}-3$$

Rewrite the inequality by using the obtained solutions.

$$\left(u-\left(\sqrt{3}-3\right)\right)\left(u-\left(-\sqrt{3}-3\right)\right)\leq 0$$

For the product to be ≤0, one of the values u-\left(\sqrt{3}-3\right) and u-\left(-\sqrt{3}-3\right) has to be ≥0 and the other has to be ≤0. Consider the case when u-\left(\sqrt{3}-3\right)\geq 0 and u-\left(-\sqrt{3}-3\right)\leq 0.

$$u-\left(\sqrt{3}-3\right)\geq 0 u-\left(-\sqrt{3}-3\right)\leq 0$$

This is false for any u.

$$u\in \emptyset$$

Consider the case when u-\left(\sqrt{3}-3\right)\leq 0 and u-\left(-\sqrt{3}-3\right)\geq 0.

$$u-\left(-\sqrt{3}-3\right)\geq 0 u-\left(\sqrt{3}-3\right)\leq 0$$

The solution satisfying both inequalities is u\in \left[-\sqrt{3}-3,\sqrt{3}-3\right].

$$u\in \begin{bmatrix}-\sqrt{3}-3,\sqrt{3}-3\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$u\in \begin{bmatrix}-\sqrt{3}-3,\sqrt{3}-3\end{bmatrix}$$