What I don't get is why intuitively we know that x cannot be negative, since x^2 is always positive, but algebraically,

x^2<x
x^2-x<0
x(x-1)<0
x<0 or x<1

^here, x<0 or x<1. This shows x can be negative. Where am I going wrong?

Target question: Is x between 0 and 1?

Statement 1: x² < x
Since we know that x² is POSITIVE, we can safely divide both sides of the inequality by x²
When we do this, we get: 1 < 1/x
If 1/x is greater than 1, we know that x must be POSITIVE (i.e., x > 0).
Since x is POSITIVE, we can safely multiply both sides of the inequality by x to get: x < 1
Combine the two inequalities to get: 0 < x < 1
In other words, x IS between 0 and 1
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x is positive
This doesn't tell us much.
It could be the case that x = 1/2, in which case, x IS between 0 and 1
Or it could be the case that x = 2, in which case, x is NOT between 0 and 1
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer:

RELATED VIDEO

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Everything is correct up to step #3.
Step #4 is incorrect.

You are taking a rule about EQUATIONS and applying it to INEQUALITIES (which we can't do)
IF it we had the EQUATION x(x-1) = 0, then we could conclude that x = 0 or x = 1

However, we can't make the same kind of conclusion when it comes to inequalities.
In fact, if we test a value like x = -1, we see that the step #3 inequality, x(x-1)<0, does not hold true. So, we can't conclude that x<0

Cheers,
Brent
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Thanks - had no idea it only applied to equations. I'm trying to find a sure-fire algebraic approach to finding the range of values we can get from these types of inequalities questions; what approach would you recommend? I can see that you're determining the sign of x first and then dividing/multiplying to "solve" the inequality. Is this the general approach I should go for with these types of questions? Are there other ways to solve inequalities?

Here's a even faster solution.

Target question: Is x between 0 and 1?

Statement 1: x² < x
For all values of x, x² will equal either zero or some positive number.
Since x is greater than x², we can be certain that x is positive.
Since x is POSITIVE, we can safely divide both sides of the inequality by x to get: x < 1
If x is positive AND x < 1, then we can be certain that x IS between 0 and 1
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x is positive
This doesn't tell us much.
It could be the case that x = 1/2, in which case, x IS between 0 and 1
Or it could be the case that x = 2, in which case, x is NOT between 0 and 1
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer:

RELATED VIDEO

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This question can be treated as a quadratic inequality question, or it can be treated as a general inequality question.
This one is pretty straightforward, so I used the latter approach.
If you're given a more complex quadratic inequality, (e.g., x² - 5x - 6 > 0), then I suggest the quadratic inequality approach (see my video below).

RELATED VIDEO

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Thanks, I actually found an easier solution that works for me - from my high school days I know what quadratic graphs look like, so I know in this case that this is a positive quadratic (x^2), and drawing out the parabola instantly tells me the range of x!

Found Bunuel's graphical explanation here for anyone interested: https://gmatclub.com/forum/if-x-is-an-i ... ml#p731476

I'm impressed that you remember how to graph parabolas from high school!! I'm sure most people can't make the same claim!

Cheers,
Brent
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Now x^2 will always be +ive, and the only region where

Statement 1 is true, is in between 0 and 1

Because higher the value of exponent lower is the value in the region of 0<x<1

Example 0.10 > 0.01,when x =0.1

Statement B doesn't tell us anything

A
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