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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

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

1. x^2<x 2. x^2-x<0 3. x(x-1)<0 4. x<0 or x<1

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

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

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?

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

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?

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).

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?

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

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!

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!

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