What is the solution to the absolute function problem?

[adjacent]

Homework Statement

$|x^2-2| \leq 1$
Find $x$

The Attempt at a Solution

$x^2-2\leq 1$
$x \leq \sqrt{3}$

and

$-(x^2-2)\leq 1$
$-x^2+2 \leq 1$
$x \geq 1$

Therefore the solution is $1 \leq x \leq \sqrt{3}$
However the book gives another solution which is $-1 \leq x \leq -\sqrt{3}$
I don't know how the latter solution is found out. How can a range be multiplied by $-1$?

 

[BruceW]

be careful when taking the square root.
$$x^2 \geq 1$$
does not imply x is positive.

 

[Ray Vickson]

Homework Statement

$|x^2-2| \leq 1$
Find $x$

The Attempt at a Solution

$x^2-2\leq 1$
$x \leq \sqrt{3}$

and

$-(x^2-2)\leq 1$
$-x^2+2 \leq 1$
$x \geq 1$

Therefore the solution is $1 \leq x \leq \sqrt{3}$
However the book gives another solution which is $-1 \leq x \leq -\sqrt{3}$
I don't know how the latter solution is found out. How can a range be multiplied by $-1$?

Draw the graph of $y = x^2-2$ over some reasonable x-range. On the same plot, draw horizontal lines $y = 1$ and $y =-1$. What does your plot show?

 

[adjacent]

be careful when taking the square root.
$$x^2 \geq 1$$
does not imply x is positive.

If that's the case, can't we write $-1 \leq x \leq \sqrt{3}$ or $1 \leq x \leq -\sqrt{3}$
Why should we make the two square roots have the same sign?

 

[adjacent]

Draw the graph of $y = x^2-2$ over some reasonable x-range. On the same plot, draw horizontal lines $y = 1$ and $y =-1$. What does your plot show?

Here it is.

I don't see anything special in it

 

[Ray Vickson]

Here it is.

I don't see anything special in it

Oh? You can't see what points on the graph $y = x^2-2$ lie between y = -1 and y = +1?

 

[adjacent]

Oh? You can't see what points on the graph $y = x^2-2$ lie between y = -1 and y = +1?

$1$ to $\sqrt{3}$ when $x>0$ and $-\sqrt{3}$ to $-1$ when $x<0$?
But I am not supposed to draw a graph.

 

[Ray Vickson]

$1$ to $\sqrt{3}$ when $x>0$ and $-\sqrt{3}$ to $-1$ when $x<0$?
But I am not supposed to draw a graph.

So, don't tell anybody. You just use the graph to get a "feel" for what is going on, and to help you keep your thinking straight. After you have used the graph like that you can throw it away. That is what we all do!

 

[statdad]

If that's the case, can't we write $-1 \leq x \leq \sqrt{3}$ or $1 \leq x \leq -\sqrt{3}$
Why should we make the two square roots have the same sign?

You should see something wrong immediately with one of these: it is really true that $1 \le -\sqrt 3$?

 

[vela]

If that's the case, can't we write $-1 \leq x \leq \sqrt{3}$ or $1 \leq x \leq -\sqrt{3}$
Why should we make the two square roots have the same sign?

It's easier if you look at it this way: You got to the point
$$1 \le x^2 \le 3.$$ Now take the square root, remembering that $\sqrt {x^2} = \lvert x \rvert$.

 

[BruceW]

If that's the case, can't we write $-1 \leq x \leq \sqrt{3}$ or $1 \leq x \leq -\sqrt{3}$
Why should we make the two square roots have the same sign?

Well, your choice at the start of your solution is an if statement.
IF $x^2 \geq 2$ then $x^2-2 \leq 1$ and a bunch of stuff follows
ELSE IF $x^2 \leq 2$ then $-x^2+2 \leq 1$ and a bunch of stuff follows
So you have to treat the two IF statements as completely separate things. You cannot carry results over from one IF statement into the other IF statement. I hope that answers your question. I wasn't totally sure what you meant.

p.s. I think the other solution given by the book should be $-\sqrt{3} \leq x \leq -1$ not $-1 \leq x \leq -\sqrt{3}$

 

[adjacent]

So, don't tell anybody. You just use the graph to get a "feel" for what is going on, and to help you keep your thinking straight. After you have used the graph like that you can throw it away. That is what we all do!

Ok

You should see something wrong immediately with one of these: it is really true that $1 \le -\sqrt 3$?

:shy: No. What about $-1 \le \sqrt 3$

It's easier if you look at it this way: You got to the point
$$1 \le x^2 \le 3.$$ Now take the square root, remembering that $\sqrt {x^2} = \lvert x \rvert$.

Oh. This seems to be a good way. Thanks

Well, your choice at the start of your solution is an if statement.
IF $x^2 \geq 2$ then $x^2-2 \leq 1$ and a bunch of stuff follows
ELSE IF $x^2 \leq 2$ then $-x^2+2 \leq 1$ and a bunch of stuff follows
So you have to treat the two IF statements as completely separate things. You cannot carry results over from one IF statement into the other IF statement. I hope that answers your question. I wasn't totally sure what you meant.

Where did $x^2 \geq 2$ come from?

p.s. I think the other solution given by the book should be $-\sqrt{3} \leq x \leq -1$ not $-1 \leq x \leq -\sqrt{3}$

Yes. It was a typo. :shy:

 

[BruceW]

Where did $x^2 \geq 2$ come from?

This was your implied "choice" when you said $|x^2-2|=x^2-2$. The other possible choice was $|x^2-2|=-x^2+2$ which occurs iff $x^2 \leq 2$. So anyway, the point I was hoping to get across is that the two cases $x^2 \geq 2$ and $x^2 \leq 2$ should be considered as separate solutions to the problem. It is not OK to carry over results from one case to the other. Also, as I said before, I am not totally sure if this was where you were having trouble anyway, but I was hoping it might be useful.