I solved it as follows:
x^2-x<0
x(x-1)<0
=> x<0 and x-1<0
=> x<0 and x>1
Is this right?
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What values of x satisfy x^2<x ?
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Anju@Gurome
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Actually, no.sri_r wrote:I solved it as follows:
x^2-x<0
x(x-1)<0
=> x<0 and x-1<0
=> x<0 and x>1
Is this right?
Unless you made a typing mistake in the first red line, you are partially correct.
And the two red lines are exclusive solution, i.e. they do not hold simultaneously.
So, a better way to put them will be : either x > 0 and (x - 1) < 0 or x < 0 and (x - 1) > 0
Now, the first solution simplifies to : x > 0 and x < 1 ---> 0 < x < 1
And, the second solution simplifies to : x > 0 and x < 1 ---> This is not possible.
So the final solution is 0 < x < 1
Anju Agarwal
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srcc25anu
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value of x that satisfy x^2 < x.
x (x - 1) < 0 or x = 0 or x = 1
on the graph we plot a parabola (for quadratic equation x^2 - x < 0) as shown with roots at 1 and 0. the graph will open up since x^2 is positive.
from the graph we see that area less than 0 lies between 0 and 1
hence for the above equation to have "< 0" or NEGATIVE values, the range of x = 0 < x < 1
had it been x^2 - X > 0 the graph would have remained the same - opening up since x^2 is positive and roots = 0 and 1. but the region selected would have been the area of the curve over x-axis i.e. x > 1 or x < 0
x (x - 1) < 0 or x = 0 or x = 1
on the graph we plot a parabola (for quadratic equation x^2 - x < 0) as shown with roots at 1 and 0. the graph will open up since x^2 is positive.
from the graph we see that area less than 0 lies between 0 and 1
hence for the above equation to have "< 0" or NEGATIVE values, the range of x = 0 < x < 1
had it been x^2 - X > 0 the graph would have remained the same - opening up since x^2 is positive and roots = 0 and 1. but the region selected would have been the area of the curve over x-axis i.e. x > 1 or x < 0
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