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Re: Blog Thread IIIa : Look Who's Blogging Now
Zingoleb:
holy shit Jimmy, that's horrible. I'm so sorry.
scarred:
Yeah, I'm kind of speechless. That kind of shit is just insane.
snalin:
Damn, man, that's bad. Condolences. Was the uncle the cousins father?
--- Quote from: Spluff on 22 Aug 2010, 05:23 ---
--- Quote from: snalin on 22 Aug 2010, 04:32 ---It's pretty simple, and I understand everything. The problem is figuring out how I can write the equations down, since there's no examples and no teachers. Which isn't much of a problem with easier problems, but for the harder ones you need to know how to write this stuff to figure out the solution. I can see that |x - 1| + |x + 1| = 2 gives that x equals (-1, 1), but I don't know how to get that down on paper.
Also what Spluff said. I think they said that we were going to use absolute values to learn about limits, but I'm not sure. Does that make any sense?
--- End quote ---
To get that down on paper you have to remember how to find the absolute value in the first place, which is √(x²) = |x|. Then you can just substitute it in, rearrange it, and then solve.
So you'd take
|x+1| + |x-1| = 2
√(x+1)² + √(x-1)² = 2
(x² + 2x + 1) + (x² - 2x + 1) = 4
x² = 1
x = ±1
--- End quote ---
But x = ±1 isn't the right answer - the correct one is (-1, 1), the interval between -1 and 1. As an example, 0,5 gives:
|0,5 + 1| + |0,5 - 1| =
|-0,5|+|1,5| =
0,5+1,5 = 2
So getting the answer x = ±1 won't be getting me a full score on the exam, since there's a pretty big difference between (-1, 1) and {-1, 1}.
pwhodges:
You're right, but that doesn't give the working...
These might help you get your head round it, though they don't show the exact same problem: here and here.
Disclaimer - I haven't done any algebra since about 1980, and never had to do anything with absolute values, even in my university course in engineering science.
What I would do is remove the absolute value bars from the equation. This can be done without squares and stuff by looking at different ranges of x values. Split the equation into two:
y = |x+1| + |x-1|
y = 2
Then to simplify the first, split into three versions for the possible signs of the expressions inside the absolute bars:
for x<-1, (x+1) and (x-1) are both negative, so you have: y = -(x+1) + -(x-1) giving y = -2x
for x>=1, (x+1) and (x-1) are both positive, so you have: y = (x+1) + (x-1) giving y = 2x
for -1<=x<1, (x+1) is positive and (x-1) is negative, so you have: y = (x+1) + -(x-1) giving y = 2
from which you can see that the equation is true for all x in the interval -1 to 1.
How you are actually expected to show it, I have no idea, though! There's probably a more systematic way of expressing it.
(edit) Oh look, this approach is described here :-)
--- Quote from: Spluff on 22 Aug 2010, 05:23 ---√(x+1)² + √(x-1)² = 2
(x² + 2x + 1) + (x² - 2x + 1) = 4
--- End quote ---
This step is wrong; you have squared each term on the left rather than the whole expression.
negative creep:
--- Quote from: Jimmy the Squid on 22 Aug 2010, 06:39 ---My cousin just shot my uncle.
I'm not sure what my reaction is supposed to be but I'm very very very sad that a nice old man is dead because my cousin, who is generally a lovely person isn't taking her fucking meds.
--- End quote ---
fuck. I'm sorry, Jimmy.
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