# Solving inequality 7 <= |9 - x^2|

1. Sep 5, 2008

### roam

Hello guys!

I need some help here...

Solve the inequality $$7 \leq \left|9-x^2\right|$$

The attempt at a solution

So, you've got two cases: $$\left|9-x^2\right| =$$

1. $$9 - x^2$$
2. $$x^2 - 9$$

For the first case $$7 \leq 9-x^2$$
$$-2 \leq -x^2$$
$$\Rightarrow \sqrt{2}\geq x$$

For the second case $$7 \leq x^2 - 9$$
$$\sqrt{16} \leq x$$
$$4 \leq x$$

Anyway, my answer is only partially correct. The actual answer has to be;
$$x \in (-\infty, -4] \cup [-\sqrt{2}, \sqrt{2}] \cup [-4, \infty)$$
I'm very curious to know how you to get that answer. Thanks.

Last edited: Sep 5, 2008
2. Sep 5, 2008

### HallsofIvy

Staff Emeritus
Re: Inequality

You can't do that. The direction of the inequality changes depending upon whether you are taking the positive or negative root. For example, if 4< 9 leads to 2< 3 but also -2> -3.

The simplest way to solve a complicated inequality is to solve the associated equation first. For example, to solve $7\le |9- x^2|$ you first solve 7= |9- x2| . As you say, that gives either 7= 9- x2 or 7= x2- 9. The first is the same as x2= 2 so $x= \pm\sqrt{2}$ and the second is the same as x2= 16 so $x= \pm 4$, just as you say.

Now the point is that a continuous function can only change from "< 7" (or any number) to "> 7" where x= 7. Those 4 points, $\pm\sqrt{2}$ and $\pm 4$ divide the real numbers into 5 intervals. Check one number in each interval. If the inequality is true for that number, it is true for every number in the interval.

3. Sep 10, 2008

### roam

Re: Inequality

Explain this please.

I appreciate it if you could explain how you suddenly get from x2≥ 16 to x≥±4
Where did you get the ± from? That's what I don't understand.

I've been doing this question over and over again and looking at similar worked examples and this sounds completely nuts to me.

I post my working again.

$$7 \leq \left|9-x^2\right|$$

By drawing the number line I find out that I need to solve the following inequalities;
i) 9-x2≤-7
ii) 9-x2≥ 7

i) 9-x2≤-7
-x2≤-16
x≥4

ii) 9-x2≥ 7
-x2≥-2
x2≤2
x≤√2

So, I got x≥4 and x≤√2 which yields; [4,∞) $$\cup$$ (-∞,√2] but I'm missing -4 and -√2 because the correct answer has to be $$x \in (-\infty, -4] \cup [-\sqrt{2}, \sqrt{2}] \cup [-4, \infty)$$.

Thanks a lot.

4. Sep 10, 2008

### snipez90

Re: Inequality

i) $$9 - x^2 \leq -7 \Rightarrow x^2 \geq 16 \Leftrightarrow |x| \geq 4 \Rightarrow x \geq 4, x \leq -4$$

The best way to understand this is to look at the equality cases in the original inequality and then look at what happens when you move past equality, i.e. when x > 4 or x < -4.

ii) $$9 - x^2 \geq 7 \Rightarrow x^2 \leq 2 \Leftrightarrow |x| \leq \sqrt{2} \Rightarrow x \leq \sqrt{2}, x \geq -\sqrt{2}$$

Again do the same type of analysis. Now ask yourself what happens when you square an inequality involving absolute value terms.

5. Sep 10, 2008

### HallsofIvy

Staff Emeritus
Re: Inequality

Where I suddenly got that? Where, in my response did I say anything like that? I wouldn't because it is not true. For example -3> -4 but (-3)2= 9< 16. What I said was that if $$x^2= 16$$ then $$x= \pm 4$$. I got the "[tex]\pm[/itex]" because (4)(4)= 16 and (-4)(-4)= 16.

Yes, you done that over and over again and everytime you do it, it is still wrong!
Taking the square root of both sides of an inequality is equivalent to dividing both sides by x- the inequality reverses depending on whether x is positive or negative and you don't know which it is!

All I can suggest is that you go back and read what I wrote again. What you shown here looks exactly what you did in your first post- and it was wrong then. You have done nothing that I suggested in my response.

6. Sep 10, 2008

### roam

Re: Inequality

I'm sorry Hall, I think I understand the error I made & I see the detail I have overlooked...
Thanks to you as well, Snipez.

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