Solve $(|x|-3)(|x|-7)\ge 0$

Question

Solve the inequality: $$(|x|-3)(|x|-7)\ge 0.$$

$$(|x|-3)(|x|-7)\ge 0 \iff \begin{cases} |x| - 3 \ge 0\\ |x|-7\ge0 \end{cases} \cup \begin{cases}|x|-3\le 0 \\ |x|-7\le 0\end{cases}.$$ Can you explain to me why this system is equivalent to $|x|\le 3 \cup |x|\ge 7.$ How to see this?

Answer 1

Let $t=|x|$, so you have to solve $$(t-3)(t-7)\geq 0$$ which is simple quadratic inequality so $t\notin (3,7)$.

Answer 2

For the first case, we have that $|x| - 7 \geq 0 \implies |x| \geq 7 \geq 3$, so: $$ |x| - 3 \geq 0 \cap |x| - 7 \geq 0 \iff |x| \geq 7 $$ Similarly, for the second case we have $|x| - 3 \leq 0 \implies |x| \leq 3 \leq 7$, so: $$ |x| - 3 \leq 0 \cap |x| - 7 \leq 0 \iff |x| \leq 3 $$ So it just reduces to $|x| \geq 7 \cup |x| \leq 3$.

Answer 3

That is because $|x|-7\ge 0 \implies |x|-3\ge 0 $ and $|x|-3\le 0 \implies |x|-7\le 0 $, so in each case, one of the inequalities is redundant.

Answer 4

Your first conditions, namely $\begin{cases} |x| - 3 \ge 0\\ |x|-7\ge 0 \end{cases} $, are saying $|x|\ge 3$ and $|x| \ge 7$. Clearly this happens if and only if $|x|\ge 7$.

By similar reasoning, your second conditions hold if and only if $|x|\le 3$.

So the union of the conditions reduces to $|x|\ge 7\cup |x|\le 3$.

Answer 5

$$\begin{cases} |x| - 3 \ge 0\\ |x|-7\ge0 \end{cases}\iff |x|\geq 7$$ and $$\begin{cases}|x|-3\le 0 \\ |x|-7\le 0\end{cases}\iff |x|\leq 3.$$

Answer 6

You have a quadratic inequality at hand. Note that if $\alpha\leq\beta$ are the (real) roots of a quadratic in $x$ whose leading coefficient is positive, then the value of the quadratic is negative when $x\in (\alpha,\beta)$. This is what has been used. Let $|x|=X$. Now you have $(X-3)(X-7)$ as your quadratic. Whenever $X$ lies between $3$ and $7$ (both non-inclusive), the quadratic must be negative. But we want it to be non-negative. Hence, you get the required result.