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Write $A \cap B$ as the union of intervals. Use standard interval notation, that may include round brackets and square brackets. None of your intervals should touch or intersect another.

$$\begin{align*} A &= \Bigl\{ x\in\mathbb{R}\Bigm| x^2 + 3x - 4 \gt 0\Bigr\}\\ B &= \Bigl\{ x\in\mathbb{R}\Bigm| x^2 + 3x - 10\leq 0\Bigr\} \end{align*}$$

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    Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and ensure that they write their answers at an appropriate level. Also, many find the use of imperative ("Prove", "Show") to be rude when asking for help; please consider rewriting your post.2012-03-07
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    Can you find another way of writing $A$? Say, as an interval or a union of interval? How about $B$? Does that help?2012-03-07
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    Do you know how to figure out what *each* of $A$ and $B$ are as (unions of) intervals? HINT: Factor the quadratics.2012-03-07

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In case this is a homework, here is a hint on a similar example.

Let $$ A = \{ x \in \mathbb{R} | x^2-x-6 > 0 \}.$$ First, factor $ x^2-x-6 = (x-3)(x+2) = 0$. Hence there are two roots at $x =3, x= -2$. This divides up the number line $\mathbb{R}$ into three intervals: $$ (-\infty, -2), (-2, 3), (3, \infty). $$ Now you need to check the sign of $(x-3)(x+2)$ in each of the three intervals. Only $x \in (3, \infty) \cup (-\infty, -2)$ makes $(x-3)(x+2)$ positive, i.e. $x^2-x-6 > 0.$ Hence $$ A = (3, \infty) \cup (-\infty, -2). $$ Do the same with $A,B$ in your question. Then $A \cup B$ should be apparent. Note: pay attention that $B$ has $\le$ in the definition, so some intervals will be closed (rather than open).

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    @ArturoMagidin Ops. I'm sorry. I did not mean to accuse the OP. I just want to help without crossing the line of `(homework)`. I'll edit my answer.2012-03-07