Solve k^2-10k+9>0

$$k ^ { 2 } - 10 k + 9 > 0$$

Solve for k

$$k\in \left(-\infty,1\right)\cup \left(9,\infty\right)$$

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.

$$k^{2}-10k+9=0$$

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, -10 for b, and 9 for c in the quadratic formula.

$$k=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$k=\frac{10±8}{2}$$

Solve the equation k=\frac{10±8}{2} when ± is plus and when ± is minus.

$$k=9 k=1$$

Rewrite the inequality by using the obtained solutions.

$$\left(k-9\right)\left(k-1\right)>0$$

For the product to be positive, k-9 and k-1 have to be both negative or both positive. Consider the case when k-9 and k-1 are both negative.

$$k-9<0 k-1<0$$

The solution satisfying both inequalities is k<1.

$$k<1$$

Consider the case when k-9 and k-1 are both positive.

$$k-1>0 k-9>0$$

The solution satisfying both inequalities is k>9.

$$k>9$$

The final solution is the union of the obtained solutions.

$$k<1\text{; }k>9$$