Solve (m+1)^2>4

$$( m + 1 ) ^ { 2 } > 4$$

Solve for m

$$m\in \left(-\infty,-3\right)\cup \left(1,\infty\right)$$

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.

$$\left(m+1\right)^{2}=0$$

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, 2 for b, and -3 for c in the quadratic formula.

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

Do the calculations.

$$m=\frac{-2±4}{2}$$

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

$$m=1 m=-3$$

Rewrite the inequality by using the obtained solutions.

$$\left(m-1\right)\left(m+3\right)>0$$

For the product to be positive, m-1 and m+3 have to be both negative or both positive. Consider the case when m-1 and m+3 are both negative.

$$m-1<0 m+3<0$$

The solution satisfying both inequalities is m<-3.

$$m<-3$$

Consider the case when m-1 and m+3 are both positive.

$$m+3>0 m-1>0$$

The solution satisfying both inequalities is m>1.

$$m>1$$

The final solution is the union of the obtained solutions.

$$m<-3\text{; }m>1$$