Proof for inequality involving absolutes

[Undirrast]

Homework Statement

If [tex]x[/tex] is real and [tex]y=\frac{x^2+4x-17}{2(x-3)}[/tex], show that [tex]|y-5| \geq 2[/tex]

Homework Equations

The Attempt at a Solution

Sorry... Absolutely no idea. I tried to substitute y into the left side to prove that [tex]-2 \leq y - 5 \leq 2[/tex] but I can't. Anything I should know to do this?

 

[JonF]

-2 <= y - 5 <= 2 would be if |y-5|<=2, which isn't the case! you got it backwards.

 

[Undirrast]

Oops... But I still can't prove it. Can't determine the range of y - 5 at all...

 

[JonF]

plug in -2 - √21 or -2 + √21

What do you get? How do you think i got those numbers? Ask yourself where is y positive? Negative? As x gets really really positive or negative what will happen to this graph? What are the possible points where y could be the smallest it could possibly be?

 

[Undirrast]

Into what? x or y? I don't get it, why should I do so?

 

[JonF]

into x

This problem might all around be easier for you if you express 5 as (10x-30)/2(x-3)

 

[Undirrast]

I got this [tex]y - 5 = \frac{(x-3)^2 - 11}{2(x-3)}[/tex], which doesn't give any minimum value when x = 3, but by splitting it up I get [tex]\frac{x-3}{2} - \frac{11}{2(x-3)}[/tex], is that what I'm supposed to do? I still don't get the 2 values you got...

 

[JonF]

turn 5 into (10x-30)/2(x-3) so you can add it to your fraction (you need like denominators). Then factor the numerator using the quadratic formula. You'll get different values than me because i didn't bother with the 5, but i suspect it will be easier for you to see if you do.

 

[Undirrast]

It's [tex]\frac{x^2 - 6x - 13}{2(x-3)}[/tex] right? Hopefully I did not calculate wrongly. If I solve [tex]x^2-6x-13=0[/tex], I'll get [tex]3 - \sqrt{22}[/tex] and [tex]3 + \sqrt{22}[/tex]. Putting them into x makes it zero, that's not what I want to prove?

 

[JonF]

Okay no more similar problems we'll just cut to the chase since I think I’m only confusing you. I’m not allowed to give the answer so I’ll try to walk you through step by step.

|(x^2 +4x -17)/2(x-3) – 5| ≥ 2 means we will have to deal with two cases
(x^2 +4x -17)/2(x-3) – 5 ≥ 2 if AND -[(x^2 +4x -17)/2(x-3) – 5] ≥ 2 Right?
Let’s deal with (x^2 +4x -17)/2(x-3) – 5 ≥ 2 first,

(x^2 +4x -17)/2(x-3) – 5 ≥ 2 ⇒ (x^2 +4x -17)/2(x-3) – 7 ≥0
do what you did on that other problem, add the -7 to (x^2 +4x -17)/2(x-3) by making like denominators then factor the numerator

 

[Undirrast]

Okay, I factorized and I got [tex](x-5)^2(x-3) \geq 0[/tex], since [tex](x-5)^2[/tex] is always positive, so I guess that leaves me to x > 3. What does this means basically?

 

[JonF]

where could this overall equation of (x-5)²/2(x-3) change from negative to positive or vice versa?

 

[Undirrast]

You mean when it's [tex]\frac{x^2-10x+25}{2(x-3)}[/tex]? When x = 5 it's zero... x < 5 is negative and x > 5 is positive...

 

[JonF]

we simplified (x^2 +4x -17)/2(x-3) – 7 ≥0

to (x-5)^2/2(x-3) ≥0 correct?

so where could switch from positive to negative?

Think of it in terms of numbers if you're confused. if we have 2 numbers: a and b when is a*a/2b negative? when is it positive?

Clearly it's when negative b is negative – do you see how this applies to this problem?

 

[Undirrast]

Sorry... I still don't see how it applies to the problem. So when x < 3 it's negative and x > 3 it's positive, then? How does this relate to the problem at all?

 

[JonF]

where is (x^2 +4x -17)/2(x-3) – 5 > 0. FYI when you did the math on this last time your -13 should be +13

Remember our goal is to prove:
|(x^2 +4x -17)/2(x-3) – 5| ≥ 2 if (x^2 +4x -17)/2(x-3) – 5 > 0



This might clear it up, let a = (x^2 +4x -17)/2(x-3) – 5.

What we're trying to do is show if a>0 then a>2
then we will show if a<0 then a>2. which will mean |a| > 2

 

[Undirrast]

Oh, okay, so how do we prove that y - 5 > 0?

 

[JonF]

We're almost half way there (don't worry the other half is the exact same process). Answer where is (x^2 +4x -17)/2(x-3) – 5 > 0. FYI when you did the math on this last time your -13 should be +13. So it's where is (x^2 - 6x + 13)/2(x-3) >0

I added this to my last post, you may have missed it, it should explain what we're trying to do:

let a = (x^2 +4x -17)/2(x-3) – 5.

What we're trying to do is show if a>0 then a>2
then we will show if a<0 then a>2. which will mean |a| > 2

 

[Undirrast]

Yea, I got the +13, so factorize it and it's [tex]\frac{(x-3)^2+4}{2(x-3)}[/tex], so it's negative when x < 3, which is unsuitable?

 

[JonF]

Great! so we know if (x^2 +4x -17)/2(x-3) – 5 > 0 then (x^2 +4x -17)/2(x-3) – 5 > 2. That's half of the inequality! How do you think we should we get the other half.

EDIT: you're not fully factoring the top, you don't want two terms if you want to find zeros. x^2 - 6x + 13 has no real roots, so it doesn't equal zero for any real x. But you're right it's negative when x<3

 

[Undirrast]

Well, [tex](x-3)^2[/tex] is definitely positive and +4 that's no doubt positive as well, doesn't seem to matter me :P

I kinda get it, but I guess it's necessary for me to explain so much text in the working as well?

 

[JonF]

You could break it down into two cases, and have almost no text.

the case of (x^2 +4x -17)/2(x-3) – 5 < 0 implies -((x^2 +4x -17)/2((x-3) – 5) > 2

and the case of (x^2 +4x -17)/2(x-3) – 5 > 0 implies (x^2 +4x -17)/2((x-3) – 5 > 2 (we just did this one)



BTW: You're right on how you dealt with (x-3)^2 + 4, I'm just half asleep at this point. I excpected you to look at the discriminate, and didn't think when I saw you didn't.

 

[Undirrast]

Thanks a lot for your tips, and your patience as well :)