Solve x^2-2x-4geq0

$$x ^ { 2 } - 2 x - 4 \geq 0$$

Solve for x

$$x\in (-\infty,1-\sqrt{5}]\cup [\sqrt{5}+1,\infty)$$

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.

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

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

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

Do the calculations.

$$x=\frac{2±2\sqrt{5}}{2}$$

Solve the equation x=\frac{2±2\sqrt{5}}{2} when ± is plus and when ± is minus.

$$x=\sqrt{5}+1 x=1-\sqrt{5}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(\sqrt{5}+1\right)\right)\left(x-\left(1-\sqrt{5}\right)\right)\geq 0$$

For the product to be ≥0, x-\left(\sqrt{5}+1\right) and x-\left(1-\sqrt{5}\right) have to be both ≤0 or both ≥0. Consider the case when x-\left(\sqrt{5}+1\right) and x-\left(1-\sqrt{5}\right) are both ≤0.

$$x-\left(\sqrt{5}+1\right)\leq 0 x-\left(1-\sqrt{5}\right)\leq 0$$

The solution satisfying both inequalities is x\leq 1-\sqrt{5}.

$$x\leq 1-\sqrt{5}$$

Consider the case when x-\left(\sqrt{5}+1\right) and x-\left(1-\sqrt{5}\right) are both ≥0.

$$x-\left(1-\sqrt{5}\right)\geq 0 x-\left(\sqrt{5}+1\right)\geq 0$$

The solution satisfying both inequalities is x\geq \sqrt{5}+1.

$$x\geq \sqrt{5}+1$$

The final solution is the union of the obtained solutions.

$$x\leq 1-\sqrt{5}\text{; }x\geq \sqrt{5}+1$$