Trying to Solve a Basic Inequality

The Student

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Solve x^2 - x < 0. We were using (a^2)^(1/2) = |a| as a method for questions like these, but I am not sure if it applies here. We also used |x| < a if and only if -a < x < a. So, I will apply both methods. x^2 - x < 0, x^2 < x, (x^2)^(1/2) < x^(1/2). So (x^2)^(1/2) < x^(1/2), |x| < x^(1/2). Then according to the methods, -x^(1/2) < x < x^(1/2). But this is as far as I can get. The answer has the solution set as (0, 1).
 
Solve x^2 - x < 0.
The related graph, y = x^2 - x, is a parabola. This is a positive quadratic, so it's an upward-opening parabola. You're wanting where the curve is below the x-axis.

The quadratic expression factors, giving you the two x-intercepts. These split the x-axis into three intervals, two where the curve is on one side of the x-axis and the interval between where the curve is on the other side of the x-axis.

Using what you know about graphing, choose the correct interval(s). ;)
 
Solve x^2 - x < 0. We were using (a^2)^(1/2) = |a| as a method for questions like. The answer has the solution set as (0, 1).

I suggest to you that there is no work involved for this question whatsoever. It is a concept question.
It says that \(\displaystyle x^2<x\), the number squared is less than itself.
That is only true on \(\displaystyle (0,1)\).
 
The related graph, y = x^2 - x, is a parabola. This is a positive quadratic, so it's an upward-opening parabola. You're wanting where the curve is below the x-axis.

The quadratic expression factors, giving you the two x-intercepts. These split the x-axis into three intervals, two where the curve is on one side of the x-axis and the interval between where the curve is on the other side of the x-axis.

Using what you know about graphing, choose the correct interval(s). ;)

Interesting, but I was only suppose to use the two methods that I mentioned if they applied.
 
I suggest to you that there is no work involved for this question whatsoever. It is a concept question.
It says that \(\displaystyle x^2<x\), the number squared is less than itself.
That is only true on \(\displaystyle (0,1)\).

Is there no method to use? How could I prove the answer?
 
This problem is a simplified version of the general case for problems like this and, although the answer to this particular question can be found easier as indicated above, I'm going to run off at the mouth (actually the hands, since I'm typing this stuff) and go the long way around.

The way to tackle problems like these is first to turn it into an equation as suggested earlier, i.e.
y = x2 - x
and then where y is zero. For the second order equation this would generally mean using the quadratic equation to find the zeros but for this particular equation we can see that the answer is x = 0 and x = 1. This means this equation can be written as
y = (x - 0) (x - 1) = x (x-1).
Now a product of two numbers is negative if one is positive and the other is negative. So do the x negative and x-1 positive first or
x < 0 and x - 1 > 0 or x < 0 and x > 1 which can't happen.

For x positive and x-1 negative we have
0 < x and x - 1 < 0 or 0 < x and x < 1 or x in the interval (0,1)

As I said, the long way around to a problem which has already been answered.
 
This problem is a simplified version of the general case for problems like this and, although the answer to this particular question can be found easier as indicated above, I'm going to run off at the mouth (actually the hands, since I'm typing this stuff) and go the long way around.

The way to tackle problems like these is first to turn it into an equation as suggested earlier, i.e.
y = x2 - x
and then where y is zero. For the second order equation this would generally mean using the quadratic equation to find the zeros but for this particular equation we can see that the answer is x = 0 and x = 1. This means this equation can be written as
y = (x - 0) (x - 1) = x (x-1).
Now a product of two numbers is negative if one is positive and the other is negative. So do the x negative and x-1 positive first or
x < 0 and x - 1 > 0 or x < 0 and x > 1 which can't happen.

For x positive and x-1 negative we have
0 < x and x - 1 < 0 or 0 < x and x < 1 or x in the interval (0,1)

As I said, the long way around to a problem which has already been answered.

Thanks for your input.
 
Interesting, but I was only suppose to use the two methods that I mentioned if they applied.
Ah. You hadn't included the instructions, and had said that you weren't sure if those methods could be applied. I hadn't realized that this meant that the instructions had stated that these were the only methods that could be applied. Sorry. :oops:

To answer your question, no, I see no way to apply only the two specified "methods" in order to solve the inequality. Please do reply with the "correct" solution, once you've learned what that is. I'd be interested to see what you had been expected to do. Thank you! ;)
 
Ah. You hadn't included the instructions, and had said that you weren't sure if those methods could be applied. I hadn't realized that this meant that the instructions had stated that these were the only methods that could be applied. Sorry. :oops:

The book gives the instructions to use the methods if we think that they are necessary, so it is quite vague about whether or not we can use that method.

To answer your question, no, I see no way to apply only the two specified "methods" in order to solve the inequality. Please do reply with the "correct" solution, once you've learned what that is. I'd be interested to see what you had been expected to do. Thank you! ;)

Unfortunately the book only gives the answer without showing any work; the answer being the solution set (0, 1).
 
\(\displaystyle x^2- x= x(x- 1)< 0\).

The product of two numbers is negative if and only if the two numbers are of different sign. That is, one is positive and the other negative.

If x> 0 and x- 1< 0 the x> 0 and x< 1 so x is any number strictly between 0 and 1. That is, 0< x< 1.

It is, of course, impossible to have x< 0 and x- 1> 0.
 
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