Solve (x+3)^2>4

$$( x + 3 ) ^ { 2 } > 4$$

Solve for x

$$x\in \left(-\infty,-5\right)\cup \left(-1,\infty\right)$$

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 5 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$x=\frac{-6±4}{2}$$

Solve the equation x=\frac{-6±4}{2} when ± is plus and when ± is minus.

$$x=-1 x=-5$$

Rewrite the inequality by using the obtained solutions.

$$\left(x+1\right)\left(x+5\right)>0$$

For the product to be positive, x+1 and x+5 have to be both negative or both positive. Consider the case when x+1 and x+5 are both negative.

$$x+1<0 x+5<0$$

The solution satisfying both inequalities is x<-5.

$$x<-5$$

Consider the case when x+1 and x+5 are both positive.

$$x+5>0 x+1>0$$

The solution satisfying both inequalities is x>-1.

$$x>-1$$

The final solution is the union of the obtained solutions.

$$x<-5\text{; }x>-1$$