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Question 813288: Solution set of: x^2 < x
with explanation please
Answer by AlbertusK(3)   (Show Source):
You can put this solution on YOUR website! Have you already learnt about inequality?
If you're still a bit confuse about this case, I will try to give some helps.
Solution set of : x^2 < x...
We can change it into x^2 - x < 0
Now we factorize them : x(x - 1) < 0
Now we should find the zero-maker. In the last calculation, we already get
x(x-10) = 0. What are the roots for this equality?
The roots are x = 0 and (x-10)=0 --> x=10.
Now we have 2 roots namely 0 and 10. CLEAR?
Now let's consider 3 cases namely :
1) All numbers less than 0,
2) All numbers between 0 and 10,
3) All numbers more than 10.
for case 1), we can try any numbers for example -1. If we substitute -1 into the first inequality from your question [x^2-x < 0], we get [1-(-1)<0] --> [2<0] and this is a contradiction.
For case 2) we can try 2. we substitute 2 and we get a contradiction again.
For case 3) we can try 12. we substitute 12 and we get a contradiction again.
So the solution set for this problem is none. (no appropriate solution set --> empty solution).
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