Solve a^2+2a-1>0

$$a ^ { 2 } + 2 a - 1 > 0$$

Solve for a

$$a\in \left(-\infty,-\sqrt{2}-1\right)\cup \left(\sqrt{2}-1,\infty\right)$$

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.

$$a^{2}+2a-1=0$$

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 -1 for c in the quadratic formula.

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

Do the calculations.

$$a=\frac{-2±2\sqrt{2}}{2}$$

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

$$a=\sqrt{2}-1 a=-\sqrt{2}-1$$

Rewrite the inequality by using the obtained solutions.

$$\left(a-\left(\sqrt{2}-1\right)\right)\left(a-\left(-\sqrt{2}-1\right)\right)>0$$

For the product to be positive, a-\left(\sqrt{2}-1\right) and a-\left(-\sqrt{2}-1\right) have to be both negative or both positive. Consider the case when a-\left(\sqrt{2}-1\right) and a-\left(-\sqrt{2}-1\right) are both negative.

$$a-\left(\sqrt{2}-1\right)<0 a-\left(-\sqrt{2}-1\right)<0$$

The solution satisfying both inequalities is a<-\left(\sqrt{2}+1\right).

$$a<-\left(\sqrt{2}+1\right)$$

Consider the case when a-\left(\sqrt{2}-1\right) and a-\left(-\sqrt{2}-1\right) are both positive.

$$a-\left(-\sqrt{2}-1\right)>0 a-\left(\sqrt{2}-1\right)>0$$

The solution satisfying both inequalities is a>\sqrt{2}-1.

$$a>\sqrt{2}-1$$

The final solution is the union of the obtained solutions.

$$a<-\sqrt{2}-1\text{; }a>\sqrt{2}-1$$