Simplifying $\{ x \in \mathbb{R} : x^2 - x - 6 \geq 0 \}$ when possible

Question

Simplify the following interval notation when possible: $$\{ x \in \mathbb{R} : x^2 - x - 6 \geq 0 \}$$

Answer 1

Hint: Find the roots (there are 2) and than think about the behaviour of $x^2-x-6$ as $|x|$ tends to infty.

Answer 2

Here is the graph, look for the points where parabola is non-negative (y ≥ 0).

Answer 3

Read about Quadratic equation. That should give you a good idea of them, then you'll be able to find the roots. According to the roots you've found, you can figure out where the equation is positive or negative.

Generally, $ax^2+bx+c$ has the opposite sign of $a$ between the roots.

The given equation has $3$ and $-2$ as roots, then where it's greater than or equal to zero?

Answer 4

This is a bit of an extension to Americo Tavares' answer.

First find any root, $x=3$ is fine. Now what you need is the second polynomial: $$ \frac{x^2 -x-6}{x-3}=Q(x) $$ where $Q(x)$ is of the form $ax+b$, the RHS becomes $$ Q(x)(x-3)=ax^2+(b-3a)x-3b $$ Then you equate coefficients and fin that $a=1,b=2$, so the original polynomial can be factored as $$ P(x)=(x-3)(x+2) \geq 0 $$ Clearly the product is larger than 0 either when bothterms are positive or when both terms are negative, hence your solution is $x \leq -2 \cup x \geq 3$