4
$\begingroup$

I have a little problem about how to wrote my answer for that question.

I know that I need to use the analyses:

$x^2 - 4 \geq 0$

$x^2 \geq 4$

$\sqrt{x^2} \geq \sqrt{4}$

$x \geq 2$

So, $[2, \infty)$

And I know that any negative value of $x^2$ became positive. And that the answer will be ($-\infty$, -2]. And the result will be $(-\infty, -2]\cup[2, \infty)$.

But how can I express that second part of the $-x$ algebrically (that word even exists? Sorry)? I only know that because it's obvious, it's logic.

But I need to express that in algebra like the positive $x$ part of the domain.

Thanka guys, cheers!

Edit:

Really thanks for all the answers. I will try every hint.

At the momment my answer became:

If $x^2 \geq a^2$ then $\|x\| \geq a$

Then:

$x^2 -4 \geq 0$

$x^2 \geq 4$

$x^2 \geq 2^2$

$\|x\| \geq 2$

i) $x \geq 2 \implies [2, \inf[$

ii) $-x \geq 2 $

$x \leq -2 \implies ]\inf, -2]$

Final answer $(-\infty, -2]\cup[2, \infty)$.

That's it I think.

Ps: Write LaTeX by mobile is a pain. :'/ Sorry for any mistakes. I can't vote up because I have not enough points. Sorry. But I appreciated all the answers. Thanks!

$\endgroup$

4 Answers 4

0
$\begingroup$

If $x^2\ge a$ then $|x|\ge \sqrt{a}$. Not $x\ge \sqrt{a}$.

Similarly, if $x^2=a$, $|x|=\sqrt{a}$. Not $x=\sqrt{a}$.

$\endgroup$
2
  • 1
    $\begingroup$ $3^2=(-3)^2$, yet $|3| \neq -3$ $\endgroup$ Sep 6, 2015 at 6:23
  • 1
    $\begingroup$ Thanks man. That solve my problem. But I don't know how to choose your answer as the solver answer through the mobile app. Or maybe I can't becaus I don't have enough points in my profile. $\endgroup$ Sep 6, 2015 at 6:30
0
$\begingroup$

HINT: $x^2\ge0$

Another HINT: You are exactly halfway to the answer

$\endgroup$
0
$\begingroup$

One slightly different way of approaching this would be to factor $x^2-4=(x+2)(x-2)$, and using the fact that the sign of a product can be determined by the sign of the factors, $$+\cdot +=+,\qquad +\cdot -=-,\qquad -\cdot -=+$$.

So, we are trying to solve when $(x+2)(x-2)>0$. Well, really, $\geq 0$, but having zero complicates the discussion of signs. So we want to know where either both factors are positive or where both factors are negative. The places where they are positive are $(-2,\infty)$ and $(2,\infty)$, and the intersection of these two intervals, the place where both are positive, is $(2,\infty)$. Similarly, they are negative on $(-\infty,-2)$ and $(-\infty,2)$, and the intersection, the place where both are negative, is $(-\infty,-2)$.

Combining these two intervals, the product is positive on both $(-\infty,-2)$ and $(2,\infty)$, and so it is positive on their union, $(-\infty,-2)\cup (2,\infty)$. Throwing in the points where the product is zero, our domain is $(-\infty,-2]\cup [2,\infty)$. Hopefully this feels slightly more systematic to you.

$\endgroup$
2
  • $\begingroup$ Nice name, dude. $\endgroup$
    – user253055
    Sep 6, 2015 at 6:09
  • $\begingroup$ @AaronThompson Thanks, but I can't really take credit for it, and I'll be damned if I'm going to let my mother take credit for it, so I will just say...thanks. $\endgroup$
    – Aaron
    Sep 6, 2015 at 6:11
0
$\begingroup$

$ x^2 - 4 ≥ 0 $. So this implies $(x+2)(x-2) ≥ 0 $ . Hence x belongs to (−∞, -2] ∪ [2, ∞).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .