Solve (x+3)^2<2

$$( x + 3 ) ^ { 2 } < 2$$

Solve for x

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

Solution Steps

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.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 6 for b, and 7 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 1\times 7}}{2}$$

Do the calculations.

$$x=\frac{-6±2\sqrt{2}}{2}$$

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

$$x=\sqrt{2}-3 x=-\sqrt{2}-3$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(\sqrt{2}-3\right)\right)\left(x-\left(-\sqrt{2}-3\right)\right)<0$$

For the product to be negative, x-\left(\sqrt{2}-3\right) and x-\left(-\sqrt{2}-3\right) have to be of the opposite signs. Consider the case when x-\left(\sqrt{2}-3\right) is positive and x-\left(-\sqrt{2}-3\right) is negative.

$$x-\left(\sqrt{2}-3\right)>0 x-\left(-\sqrt{2}-3\right)<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\left(-\sqrt{2}-3\right) is positive and x-\left(\sqrt{2}-3\right) is negative.

$$x-\left(-\sqrt{2}-3\right)>0 x-\left(\sqrt{2}-3\right)<0$$

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

$$x\in \left(-\left(\sqrt{2}+3\right),\sqrt{2}-3\right)$$

The final solution is the union of the obtained solutions.

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