How do I solve an inequality with a quadratic function?

[tmt1]

Hello,

I have this inequality:

$$-x^2 + 4 < 0$$

Then,

I get to

$$-(x-2)(x+2) < 0$$

Now, how do I solve this question from here.

I understand that x = -2, or x =2 but how do I use this to solve the inequality?

Thanks

 

[mathbalarka]

Why not simplify by chucking out the $-$ sign first?

$$-x^2 + 4 < 0 \Longrightarrow -(x^2 - 4) < 0 \Longrightarrow x^2 - 4 > 0$$

Thus, $(x - 2)(x + 2) > 0$. Let's see, try a different approach this time. The expression vanishes at $x = 2$ and $x = -2$, so divide the interval $x \in [-\infty, \infty]$ up in the intervals $[-\infty, -2) \cup [-2, 2] \cup (2, \infty]$. Note that when $x \in [-2, 2]$, $(x - 2)(x + 2) \leq 0$. Can you determine the solutions then?

 

[tmt1]

So the answer is ,

$$x \ge -2 $$ or $$x \le 2$$ ?

By the way , what is $$\in$$ so I can find more information about it?

Thanks

 

[mathbalarka]

So the answer is ,

$$x \ge -2 $$ or $$x \le 2$$ ?

Unfortunately, no, that's not it. For example, pick up $x = 1$ (note that $1 \leq 2$ and $1 \geq - 2$). $-x^2 + 4 = -1 + 4 = 3$ which is NOT $< 0$.

By the way , what is $$\in$$ so I can find more information about it

It's the "belongs to" notation. If you're not familiar with intervals, I am afraid you'd have to do it in the usual way I've mentioned in similar questions before.

 

[tmt1]

Unfortunately, no, that's not it. For example, pick up $x = 1$ (note that $1 \leq 2$ and $1 \geq - 2$). $-x^2 + 4 = -1 + 4 = 3$ which is NOT $< 0$.
It's the "belongs to" notation. If you're not familiar with intervals, I am afraid you'd have to do it in the usual way I've mentioned in similar questions before.

Okay, so it is the opposite of what I said (I think):

$$x \le -2 $$ or $$x \ge 2$$

 

[mathbalarka]

That is indeed that case (Yes) but can you explain why?

 

[tmt1]

That is indeed that case (Yes) but can you explain why?

Well, I can in this case, as you pointed out, anything between -2 and 2 would not solve the inequality. Whereas, less than -2 would fulfill the inequality as would greater than 2, they both share the same result since $$-x^2 = -(-x)^2$$

 

[mathbalarka]

Yes, well, in general, if any function $f(x)$ has a root (a solution to $f(x) = 0$) at $x = x_0$, then $f(x)$ "changes sign" in the neighborhood of $x = x_0$ , i.e., $\text{sgn}(f(x))$ changes from $1$ to $-1$ or $-1$ to $1$ as $x$ moves from $a$ to $b$ for some sufficiently small interval $(a, b)$ containing $x_0$.

That is the fact I used for $f(x) = x^2 - 4$, noting that the only roots occur at $x = -2$ and $x = 2$.