Solve m^2+4m-4<0

$$m ^ { 2 } + 4 m - 4 < 0$$

Solve for m

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

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.

$$m^{2}+4m-4=0$$

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, 4 for b, and -4 for c in the quadratic formula.

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

Do the calculations.

$$m=\frac{-4±4\sqrt{2}}{2}$$

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

$$m=2\sqrt{2}-2 m=-2\sqrt{2}-2$$

Rewrite the inequality by using the obtained solutions.

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

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

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

This is false for any m.

$$m\in \emptyset$$

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

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

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

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

The final solution is the union of the obtained solutions.

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