Solve n^2-11n+5>0

$$n ^ { 2 } - 11 n + 5 > 0$$

Solve for n

$$n\in \left(-\infty,\frac{11-\sqrt{101}}{2}\right)\cup \left(\frac{\sqrt{101}+11}{2},\infty\right)$$

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.

$$n^{2}-11n+5=0$$

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, -11 for b, and 5 for c in the quadratic formula.

$$n=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$n=\frac{11±\sqrt{101}}{2}$$

Solve the equation n=\frac{11±\sqrt{101}}{2} when ± is plus and when ± is minus.

$$n=\frac{\sqrt{101}+11}{2} n=\frac{11-\sqrt{101}}{2}$$

Rewrite the inequality by using the obtained solutions.

$$\left(n-\frac{\sqrt{101}+11}{2}\right)\left(n-\frac{11-\sqrt{101}}{2}\right)>0$$

For the product to be positive, n-\frac{\sqrt{101}+11}{2} and n-\frac{11-\sqrt{101}}{2} have to be both negative or both positive. Consider the case when n-\frac{\sqrt{101}+11}{2} and n-\frac{11-\sqrt{101}}{2} are both negative.

$$n-\frac{\sqrt{101}+11}{2}<0 n-\frac{11-\sqrt{101}}{2}<0$$

The solution satisfying both inequalities is n<\frac{11-\sqrt{101}}{2}.

$$n<\frac{11-\sqrt{101}}{2}$$

Consider the case when n-\frac{\sqrt{101}+11}{2} and n-\frac{11-\sqrt{101}}{2} are both positive.

$$n-\frac{11-\sqrt{101}}{2}>0 n-\frac{\sqrt{101}+11}{2}>0$$

The solution satisfying both inequalities is n>\frac{\sqrt{101}+11}{2}.

$$n>\frac{\sqrt{101}+11}{2}$$

The final solution is the union of the obtained solutions.

$$n<\frac{11-\sqrt{101}}{2}\text{; }n>\frac{\sqrt{101}+11}{2}$$