very sorry for the noobish post, I'm studying math on my own and lot of times im very confused about little things.
Here is my question, the original problem starts out like this: Solve the inequality $\vert x -2 \vert \geq \vert 2x -3\vert $
Here is the solved problem from my book
- $\vert x -2 \vert \geq \vert 2x -3\vert $
- $ (x-2)^2 \geq (2x-3)^2 $
- $ 3x^2 - 8x + 5 \leq 0 $
- $ (x-1)(3x-5) \leq 0$
- $1 \leq x \leq 5/3$.
How does: $(x-1)(3x-5) \leq 0$. (Step 4)
Become solved as : $1 \leq x \leq 5/3$. (Step 5)
If it were separate wouldn't it be:
$$ x \leq 1 \qquad \text{and} \qquad x \leq 5/3 $$
So how does $x\leq 1$ become $1\leq x$??
Thanks so much!! =)
So you want $(x-1) (3x-5) \leq 0$. So either one of the terms has to be positive and the other has to be negative.
So if $x \geq 5/3$, then both the terms are positive and the product is positive.
Similarly if $x \leq 1$, then both the terms are negative and their product is positive.
So observe that you don't want either of these things to happen, that is $x \notin (- \infty,1) \cup (5/3,\infty)$