Absolute Value question. (1 Viewer)

[coroneos]

What do you think is the most efficient and ultimately the fastest way of solving the following:

|x-8| < |2x+8|

i say graphically. What do you think?

 

[flyin']

Yes, I agree - graphically, and then calculate the points where they cut into each other.

 

[...]

but graphically, would they get u the marks for the working out???

 

[ezzy85]

i solve x-8 < 2x + 8 and then x-8 > -2x-8
its always worked in exams.

 

[iambored]

fastest would be graphically, however i don't think u'd get marks for working

 

[flyin']

It very much depends on the amount of time you have left. By the time you reach these in the Exam room, you might not have the luxury of workin' it out by properly.

 

[...]

but usually absolute q's are one of the early ones....

 

[Fosweb]

As if you wouldnt get marks for working. Unless it says: solve algebraically, then I would do these types of questions graphically.

 

[flyin']

Well, the early Inequality (Ext1) questions are not of the standard as the first post in this thread. You can usually just factorise and graph them, and the solution is obvious.

 

[Rahul]

another method,

|x-8| < |2x+8|
(x-8)^2 < (2x+8)^2
...and then solve

this method is probably better for more complex functions that include |x^3+4x^2+2| or something of that sort.

but other than that, for one like this one, graphing it would be a good method. of course you would have to sketch both |x-8| and |2x+8| and then simultaneously solve both lines, which will be different(2 sets of different simultaneous equations)

 

[...]

Originally posted by Rahul
another method,

|x-8| < |2x+8|
(x-8)^2 < (2x+8)^2
...and then solve

yea, this is probably the method best to use...

 

[flyin']

I remember you have to be careful with that method. Check some examples and some Exam reports, I say.

 

[coroneos]

Originally posted by ...
yea, this is probably the method best to use...

The problem with that is you have to make sure you
a) do not stuff up any negative signs... equality signs etc.
b) - more importantly - you get 2 values usually and one never seems to work, despite the fact that it works graphically after collecting like terms etc..

weird..

I don't know. My teacher told me to do it graphically.

 

[wogboy]

I personally have to agree with ezzy85, anything in the form: |a| < |b|
can be rewritten as two inequations:
a < b & a > -b (which is actually equivalent to -b < a < b)

Just two simple linear inequalities to solve, no quadratics or graphing inconveniences (just remember it is the INTERSECTION of these two solutions which gives the final solution not the UNION, don't get this confused).

The other method with quadratics (squaring the absolute value) is better for the more complicated questions though,

 

[JOhnNiLiCiouS]

Originally posted by iambored
fastest would be graphically

I kinda disagree. If you drew graphs, you still have to work out when they meet... ie algebraically. But if your confident in your graphing and can draw perfectly 45 degree lines... and considering that x value DON'T have to be whole numbers... I'd strongly suggest do it Rahul's way.

STRONGLY... but unless its like a 1 mark question and its heaps obvious . In our trial there was one and it was worth 4 marks...

 

[fatmuscle]

JOhnNiLiCiouS, u ain't shelli's bro are you?

 

[Viper]

i solved it saying LHS = RHS and LHS = -RHS.

The answer i got was: x<=-16, x>=0.

Does this look right?

Cheers

 

[:: ck ::]

edit : wait its not correct...
didnt look at ur answer properly...

why do u have a greater/equal sign and a less than/equal sign

 

[jogloran]

Are you sure? I did it by using |x| = sqrt(x^2) (squaring both sides) and got x < 0, x > 16/3. I used a computer program to check it over and I got the same result.

(x-8)^2 < (2x-8)^2
x^2 - 16x + 63 < 4x^2 - 32x + 63
-3x^2 + 16x < 0
-3x (x-16/3) < 0
x < 0, x > 16/3

 

[absolution*]

8^2 = 63?

I got the same as viper without the "equal to" signs.