Inequality

**RedBarchetta** · November 23rd 2008, 01:51 AM

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

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

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

**mr fantastic** · November 23rd 2008, 01:56 AM

Originally Posted by RedBarchetta

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

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

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

Try drawing the graphs of $y = |x^2 - 1|$ and $y = 2$ . Now calculate the x-coordinate of the intersection points of these two graphs:

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

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

**RedBarchetta** · November 23rd 2008, 02:05 AM

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.

**Moo** · November 23rd 2008, 02:24 AM

Hello,

Why doesn't it work here ?

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

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

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

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

~~~~~~~~~~~~~~~~~~~~~~
note : $|a|0, $x^2<a^2 \Leftrightarrow |x|<a$

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

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

**RedBarchetta** · November 23rd 2008, 09:13 AM

Thanks guys.

Thread: Inequality

LinkBack

Thread Tools

Display

Inequality

Similar Math Help Forum Discussions

Inequality with x in the denominator (rational inequality)

Proving the triangle inequality using the Cauchy-Schwartzs inequality

inequality

Inequality help

Inequality :\

Search Tags