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16 Jul 2020, 09:41
Kudos
Bookmarks
Hi,
Why can't I look at this absolute equation this way.
x|x| - x < 0
x [|x| - 1] <0
Therefore 1) x is <0 or 2) |x|-1 is <0
From 2):
|x|<1
x can be negative fraction or positive fraction.
However when I test a negative fraction back to the main equation, it does not work.
My question is why can we not solve an absolute equation this way?
If so, we have to approach absolute equations mainly using logic or testing cases since conventional algebra does not work in this case?
Why can't I look at this absolute equation this way.
x|x| - x < 0
x [|x| - 1] <0
Therefore 1) x is <0 or 2) |x|-1 is <0
From 2):
|x|<1
x can be negative fraction or positive fraction.
However when I test a negative fraction back to the main equation, it does not work.
My question is why can we not solve an absolute equation this way?
If so, we have to approach absolute equations mainly using logic or testing cases since conventional algebra does not work in this case?
17 Jul 2020, 05:16
Kudos
Bookmarks
weschan90
You can still solve it this way, no issues here:
x [|x| - 1] <0
After this step, always look at the inflection points: -1, 0, 1
Now make ranges:
x < -1
-1 < x < 0
0 < x < 1
x > 1
And then see which ranges follow the inequality by substituting values from that range. This is a foolproof method which will work every time without any confusion!
Try substituting -2, -0.5, 0.5 and 2
We see first and third range follows the inequality.
Drop a kudos if this helps!
General Discussion
16 Jul 2020, 10:48
Kudos
Bookmarks
weschan90
x|x| - x < 0
x (|x| - 1) < 0
In the two entities 'x' and '|x|-1' either can be negative and when other is positive.
Case 1: x < 0 and |x|-1 > 0
Thus, |x| > 1 => x < -1 or x > 1
But on checking you can find that for x = 2 and x = -1/2, x|x| - 1 < 0 in invalidated.
Hence x < -1
Case 2: x > 0 and |x| - 1 < 0
Thus, |x| < 1 which is only possible when -1 < x < 1 as |x| > 0
Hence 0 < x < 1
Test numbers if you want.
