Solve 11x^2-9x+1>0

$$11 { x }^{ 2 } -9x+1 > 0$$

Solve for x

$$x\in \left(-\infty,\frac{9-\sqrt{37}}{22}\right)\cup \left(\frac{\sqrt{37}+9}{22},\infty\right)$$

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.

$$11x^{2}-9x+1=0$$

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 11 for a, -9 for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 11\times 1}}{2\times 11}$$

Do the calculations.

$$x=\frac{9±\sqrt{37}}{22}$$

Solve the equation x=\frac{9±\sqrt{37}}{22} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{37}+9}{22} x=\frac{9-\sqrt{37}}{22}$$

Rewrite the inequality by using the obtained solutions.

$$11\left(x-\frac{\sqrt{37}+9}{22}\right)\left(x-\frac{9-\sqrt{37}}{22}\right)>0$$

For the product to be positive, x-\frac{\sqrt{37}+9}{22} and x-\frac{9-\sqrt{37}}{22} have to be both negative or both positive. Consider the case when x-\frac{\sqrt{37}+9}{22} and x-\frac{9-\sqrt{37}}{22} are both negative.

$$x-\frac{\sqrt{37}+9}{22}<0 x-\frac{9-\sqrt{37}}{22}<0$$

The solution satisfying both inequalities is x<\frac{9-\sqrt{37}}{22}.

$$x<\frac{9-\sqrt{37}}{22}$$

Consider the case when x-\frac{\sqrt{37}+9}{22} and x-\frac{9-\sqrt{37}}{22} are both positive.

$$x-\frac{9-\sqrt{37}}{22}>0 x-\frac{\sqrt{37}+9}{22}>0$$

The solution satisfying both inequalities is x>\frac{\sqrt{37}+9}{22}.

$$x>\frac{\sqrt{37}+9}{22}$$

The final solution is the union of the obtained solutions.

$$x<\frac{9-\sqrt{37}}{22}\text{; }x>\frac{\sqrt{37}+9}{22}$$