1. ## Inequality

Can anyone tell me how my book went from point a to b?

$\displaystyle |x^2 - 1| < 2 \Rightarrow - \sqrt 3 < x < \sqrt 3$

I just don't see it...thanks.

2. Originally Posted by RedBarchetta
Can anyone tell me how my book went from point a to b?

$\displaystyle |x^2 - 1| < 2 \Rightarrow - \sqrt 3 < x < \sqrt 3$

I just don't see it...thanks.
Try drawing the graphs of $\displaystyle y = |x^2 - 1|$ and $\displaystyle y = 2$. Now calculate the x-coordinate of the intersection points of these two graphs:

$\displaystyle x^2 - 1 = 2 \Rightarrow x^2 = 3 \Rightarrow x = \sqrt{3}$ or $\displaystyle x = -\sqrt{3}$.

For what values of x is the graph of $\displaystyle y = |x^2 - 1|$ below the graph of $\displaystyle y = 2$ ....?

3. Ah ha, now that makes much more sense graphically.

How would you solve this without looking at it graphically?

Using the typical method for solving an inequality doesn't work here.

4. Hello,

Why doesn't it work here ?

$\displaystyle |x^2-1|<2 \Leftrightarrow -2<x^2-1<2 \Leftrightarrow -1<x^2<3$
Whatever x is, the latter inequality can be restricted to :
$\displaystyle x^2<3$, since $\displaystyle x^2 \geq 0$.

So we have this new inequality :
$\displaystyle x^2<3$

which gives $\displaystyle |x|< \sqrt{3}$ (*)

that is to say $\displaystyle -\sqrt{3}<x<\sqrt{3}$

~~~~~~~~~~~~~~~~~~~~~~
note : $\displaystyle |a|<b \Leftrightarrow -b<a<b$, where b is positive.

(*) this can be proved.
a>0, $\displaystyle x^2<a^2 \Leftrightarrow |x|<a$

$\displaystyle x^2-a^2<0$
$\displaystyle (x-a)(x+a)<0$

so either $\displaystyle x-a<0$ and $\displaystyle x+a>0$ ----> x < a and x > -a. This is equivalent to |x|< a
either $\displaystyle x-a>0$ and $\displaystyle x+a<0$ -----> x > a and x < -a. This is not possible since a > 0.

5. Thanks guys.