Is x^2 greater than x ?

Thank you both!

It seems that I just memorized the method and didn't analyze it well.

I think that I get it now. And also I think that I have found another way that also can be helpful to you. Please, confirm if I am Ok. Please, Bunuel, take a look too.


(1) \(x^2 > x^3\).
\(x^2 - x^3 > 0\)
\(x^2 * (1-x) > 0\)
So, we get: x= 0 and x = 1

If \(x>1\) -----> \(x^2 * (1-x) < 0\)
If \(0<x<1\) -----> \(x^2 * (1-x) > 0\)
If \(x<0\) -----> \(x^2 * (1-x) > 0\)

So, we get:

-----(+)------0----(+)-------1------(-)------

Therefore, the solution is x < 1.

The same solution using this method:
\(x^2 > x^3\)
We divide by \(x^2\)
\(1 > x\)
\(x < 1\).

What do you think?

Hi,

As you can see, the sign only depends on 1-x, so, even powers can be ignored here.

Regards,

Is x^2 greater than x ?

Question: is \(x^2>x\)? --> is \(x(x-1)>0\)? is \(x<0\) or \(x>1\)?

(1) x^2 > x^3. Now, from this statement we know that \(x\neq{0}\), so \(x^2>0\) and we can safely reduce by it to get \(1>x\) (\(x<1\)). So, finally we have that given inequality holds true for \(x<0\) and \(0<x<1\) (remember we should exclude 0 from the range, since if \(x=0\) then the given inequality doesn't hold true). Not sufficient.

(2) x^2 > x^4. Apply the same logic here: we know that \(x\neq{0}\), so \(x^2>0\) and we can safely reduce by it to get \(1>x^2\) (\(x^2<1\)) --> \(-1<x<1\). So, finally we have that given inequality holds true for \(-1<x<0\) and \(0<x<1\) (remember we should exclude 0 from the range, since if \(x=0\) then the given inequality doesn't hold true). Not sufficient.Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) gives: \(-1<x<0\) and \(0<x<1\), so we can still have an YES answer (consider \(x=-0.5\)) as well as a NO answer (consider \(x=0.5\)).

Answer: E.

Hope it's clear.

The red parts are not correct. You should exclude zero, from all the ranges, since if \(x=0\), then neither of statements hold true.

Hope it's clear.

Hi,

Thanks for pointing it out!

Regards,

Just one question about the highlighted part of the explanation: It is clear that \(x^2\) is always positive. However, is the division by \(x^2\) a valid inequality operation without knowing whether x is positive or negative?

When dividing/multiplying an inequality by a variable we need to know its sign. If it's positive we should keep the sign and if its negative we should flip the sign. That's the rule. We know that x^2 is positive, so we can safely multiply/divide an inequality by it (so in this case it does not matter whether x itself is positive or negative).

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