How Do You Solve Complex Inequalities Involving Absolute Values and Ratios?

[sacwchiri]

ok so i missed a few classes and i have a test comming up soo and i need some help understanding inequalities.

the example I am trying to do is:

If |x-2|< 2, show that |(x^2+3x+1)/(x+1)| < 7

form what i could understand from a textbook was something like:

-2<x-2<2
0 < x < 4

and from there on i do a restriction for the x+1 so it can't be divided by 0 right and after that i just kinda get lost... i think i have to find when the top one becomes 0 too but its not exact and i would have to use -b+-sqrt() equation thing soo any help on this one really helps

 

[arildno]

Try first to find:

1. The x-values solving:
\frac{x^{2}+3x+1}{x+1}=7

and thereafter,
2. The x-values solving:
\frac{x^{2}+3x+1}{x+1}=-7

 

[sacwchiri]

so if that is that and i have a restriction on the lower value i can multiply without reversing the sign right?? soo
x^2 + 3x + 1 = 7(x+1)
X^2 - 4x -6 = 0

and then use -b formula thing
a = 1, b = -4, c = -6

(4+-sqrt(4^2 - 4(1)(-6)))/2
(4+- sqrt (40))/2
? for some reason it sound like I am doing something wrong

 

[arildno]

Why should you be wrong?

Let's see what we get out of this!

Your two roots are
2\pm\sqrt{10}

Note that your interval [0,4] from your first equality lies strictly between these two numbers.

Furthermore, for any particular non-negative number, say 0 you choose between your two roots, the fraction is LESS than 7.
Thus, for ALL such numbers lying between your two roots, the fraction must be less than 7, since the fraction is continuous in x.

But that means, in particular, that if x lies in [0,4], then the fraction is less than 7, which was to be proven.

 

[sacwchiri]

wait ... where did that

2\pm\sqrt{10}

edit: i found i see where you got it but still... I am lost..
the original range for x was ]0,4[...
and one more thing I am seeing here that there is a table in which you put this values to see if they are positive or negative once evaluated... so maybe i could use that?