Solve a^2-4a+1>0

$$a ^ { 2 } - 4 a + 1 > 0$$

Solve for a

$$a\in \left(-\infty,2-\sqrt{3}\right)\cup \left(\sqrt{3}+2,\infty\right)$$

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.

$$a^{2}-4a+1=0$$

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, -4 for b, and 1 for c in the quadratic formula.

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

Do the calculations.

$$a=\frac{4±2\sqrt{3}}{2}$$

Solve the equation a=\frac{4±2\sqrt{3}}{2} when ± is plus and when ± is minus.

$$a=\sqrt{3}+2 a=2-\sqrt{3}$$

Rewrite the inequality by using the obtained solutions.

$$\left(a-\left(\sqrt{3}+2\right)\right)\left(a-\left(2-\sqrt{3}\right)\right)>0$$

For the product to be positive, a-\left(\sqrt{3}+2\right) and a-\left(2-\sqrt{3}\right) have to be both negative or both positive. Consider the case when a-\left(\sqrt{3}+2\right) and a-\left(2-\sqrt{3}\right) are both negative.

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

The solution satisfying both inequalities is a<2-\sqrt{3}.

$$a<2-\sqrt{3}$$

Consider the case when a-\left(\sqrt{3}+2\right) and a-\left(2-\sqrt{3}\right) are both positive.

$$a-\left(2-\sqrt{3}\right)>0 a-\left(\sqrt{3}+2\right)>0$$

The solution satisfying both inequalities is a>\sqrt{3}+2.

$$a>\sqrt{3}+2$$

The final solution is the union of the obtained solutions.

$$a<2-\sqrt{3}\text{; }a>\sqrt{3}+2$$