Solve 2b^2-4b+1>0

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

Solve for b

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

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.

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

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

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

Do the calculations.

$$b=\frac{4±2\sqrt{2}}{4}$$

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

$$b=\frac{\sqrt{2}}{2}+1 b=-\frac{\sqrt{2}}{2}+1$$

Rewrite the inequality by using the obtained solutions.

$$2\left(b-\left(\frac{\sqrt{2}}{2}+1\right)\right)\left(b-\left(-\frac{\sqrt{2}}{2}+1\right)\right)>0$$

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

$$b-\left(\frac{\sqrt{2}}{2}+1\right)<0 b-\left(-\frac{\sqrt{2}}{2}+1\right)<0$$

The solution satisfying both inequalities is b<-\frac{\sqrt{2}}{2}+1.

$$b<-\frac{\sqrt{2}}{2}+1$$

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

$$b-\left(-\frac{\sqrt{2}}{2}+1\right)>0 b-\left(\frac{\sqrt{2}}{2}+1\right)>0$$

The solution satisfying both inequalities is b>\frac{\sqrt{2}}{2}+1.

$$b>\frac{\sqrt{2}}{2}+1$$

The final solution is the union of the obtained solutions.

$$b<-\frac{\sqrt{2}}{2}+1\text{; }b>\frac{\sqrt{2}}{2}+1$$