Solve k^2-k-4>0

$$k ^ { 2 } - k - 4 > 0$$

Solve for k

$$k\in \left(-\infty,\frac{1-\sqrt{17}}{2}\right)\cup \left(\frac{\sqrt{17}+1}{2},\infty\right)$$

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.

$$k^{2}-k-4=0$$

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

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

Do the calculations.

$$k=\frac{1±\sqrt{17}}{2}$$

Solve the equation k=\frac{1±\sqrt{17}}{2} when ± is plus and when ± is minus.

$$k=\frac{\sqrt{17}+1}{2} k=\frac{1-\sqrt{17}}{2}$$

Rewrite the inequality by using the obtained solutions.

$$\left(k-\frac{\sqrt{17}+1}{2}\right)\left(k-\frac{1-\sqrt{17}}{2}\right)>0$$

For the product to be positive, k-\frac{\sqrt{17}+1}{2} and k-\frac{1-\sqrt{17}}{2} have to be both negative or both positive. Consider the case when k-\frac{\sqrt{17}+1}{2} and k-\frac{1-\sqrt{17}}{2} are both negative.

$$k-\frac{\sqrt{17}+1}{2}<0 k-\frac{1-\sqrt{17}}{2}<0$$

The solution satisfying both inequalities is k<\frac{1-\sqrt{17}}{2}.

$$k<\frac{1-\sqrt{17}}{2}$$

Consider the case when k-\frac{\sqrt{17}+1}{2} and k-\frac{1-\sqrt{17}}{2} are both positive.

$$k-\frac{1-\sqrt{17}}{2}>0 k-\frac{\sqrt{17}+1}{2}>0$$

The solution satisfying both inequalities is k>\frac{\sqrt{17}+1}{2}.

$$k>\frac{\sqrt{17}+1}{2}$$

The final solution is the union of the obtained solutions.

$$k<\frac{1-\sqrt{17}}{2}\text{; }k>\frac{\sqrt{17}+1}{2}$$