x^2 -x <0
x^2 - x < 0
x^2 < x
x < sqrt(x)
x must be greater than 0 since sqrt(x) is imaginary for negative x, and 0 < 0 is not true.
Visualizing the graphs of x and sqrt x, one can imediately see that x < sqrt(x) in the range 0 < x < 1. To see this, consider that 1 = sqrt(1), therefore the range will be either greater or less than 1.
0.5 < sqrt(0.5) = .707
4 > sqrt(4) = 2
Therefore the range is (0,1)
x^2 -x < 0
Let us factor x:
x(x-1) < 0
In order for the inequality to hold, we have two options:
x < 0 and x-1 >0
x<0 and x > 1
This is impossible becayse x can not be <0 and >1 at the same time.
The other option is:
x>0 and x-1 <0
x>0 and x<1
Then x belongs to the interval (0,1)
First of all, let's solve the expression like an equation. For this reason, we'll put:
x^2 -x = 0
We'll factorize and we'll get:
x(x-1) = 0
We'll put each factor equal to zero.
x = 0
x-1 = 0
x = 1
Now, we'll follow the rule: between the solution of the equation, the expression will have the opposite sign of the coefficient of x^2, that means that x(x-1)<0, outside the solutions, the expression will be positive.
So, the conclusion is that the expression is negative for values of x which are in the interval (0,1).
To solve x^2-x < 0
We factorise the left and then the inequation is:
x(x-1) < 0.
So to have the negative values the values of should belong to the interval (0,1) where 0 and 1 are the rooots of x(x-1) = 0.
So the solution set of x belongs to (0,1) for which x^2-x < 0.