0
$\begingroup$

The question is very simple but my brain forgot some theory for 5-6 years. Please help me

I have this simple |4x-2| >= -1

Sorry if I took your time for explanation.

Thank you

  • 0
    @AlexBecker: True, sorry for my bad english2012-09-25

3 Answers 3

1

All $x$ numbers are solutions, since $|z| \ge 0$ for any (real or complex) number, and $0\ge -1$ and $\ge$ is transitive.

0

I think 'module' is a mistranslation into English from your native language. This is an inequality which is true for any $x$, because absolute value always returns a nonnegative number.

  • 0
    Sure. But $x$ can be any complex number and your inequality will still be satisfied. There are infinitely many solutions.2012-09-25
0

Here is an example on how to solve it, in the case where not all $x$'s are solutions. Say you had to solve $x$ such that $ |4x-2|\leq 1. $ What $| x |\leq n$ means is that $x$ is comprised between $-n$ and $n$, and so you have $-n\leq x\leq n$. In your (my) case, it would become \begin{array}{rll} -1\leq& 4x-2&\leq 1 \\ 1\leq& 4x&\leq 3\\ \frac{1}{4}\leq& x&\leq \frac{3}{4} \end{array} so for $\geq$ inequality, you take the complement of that set in $\mathbb{R}$.

  • 0
    for some reason I did the case $|4x-2|\leq 1$2012-09-25