Solution set of modulus inequations $$|2x+5|\leq\dfrac{1}{2}$$
What will be the solution set?
My attempt:


*

*For $x \in \left(-\infty,-\dfrac{5}{2}\right)$
$x\geq -\dfrac{11}{4}$

*For $x \in \left[-\dfrac{5}{2},\infty\right)$
$x\leq -\dfrac{9}{4}$
Now do we have to take union or intersection? How do we find what to do? In this case we have to take intersection, but why?

 A: For real $x,$
$$|2x+5|\le\dfrac12\iff(2x+5)^2\le\left(\dfrac12\right)^2$$
$$\iff(4x+10)^2\le1\iff(4x+10-1)(4x+10+1)\le0$$
$$\iff\left[x-\left(-\dfrac94\right)\right]\cdot\left[x-\left(-\dfrac{11}4\right)\right]\le0$$
Now if $(x-a)(x-b)\le0$ where $a<b;$ we can prove $a\le x\le b$
A: Because:
$$|2x+5|\leq\dfrac{1}{2} \iff -\dfrac{1}{2}\leq 2x+5\leq\dfrac{1}{2} \\
\iff -\dfrac{1}{2}-5\leq 2x\leq\dfrac{1}{2}-5 \iff \frac{-11}{14}\leq x\leq \frac{9}{4}$$
which means that $\frac{-11}{14}\leq x$ and $x\leq \frac{9}{4}$, so you have to take the intersection, because you have to satisfy both inequatilities simultaneously.
A: rewrite you inequality as $$|x-(-5/2)| \le \frac 14.  \tag 1$$ and interpret the left hand side as the distance between the points $x$ and $-\frac12$ on the ncumber line. so what you are looking for is all points $x$ that is within a distance of $\frac 14$ from the points $-\frac 52.$  they are the points $x$ between $-\frac{11}4$ and $-\frac 94.$ that is $$-\frac{11}4 \le x \le -\frac 94. $$
A: For example, if $$|x|>\dfrac{1}{2}$$ then $$+x>\dfrac{1}{2} \vee x<-\dfrac{1}{2}$$
Here is the trick, the $\vee$ reminds you to take the union. $\vee$ and $\cup$, $\wedge$ and $\cap$
