Is x x^2 ? (1) x is greater than x^3 (2) x is greater t

Question

Is x > x^2 ?

(1) x is greater than x^3

(2) x is greater than x^4

hi can you solve this Agebraically instead of picking numbers

Bunuel · Answer

Is x > x^2 --> is x(x-1)<0 ? --> is 0 x^3 --> x*(x^2-1)<0 --> (x+1)*x*(x-1)<0 --> 0 x^4 --> 0is-x-0-x-x-2-x-x-3-i-did-this-algebraically-i-108110.html is-x-between-0-and-1-1-x-x-3-2-x-x-107401.html is-x-0-s1-x-x-2-s2-x-2-x-108207.html is-x-x-3-s1-x-0-s2-x-2-x-108395.html Check this for more on solving this kind of inequalities: x2-4x-94661.html#p731476 inequalities-trick-91482.html everything-is-less-than-zero-108884.html?hilit=extreme#p868863 xy-plane-71492.html?hilit=solving%20quadratic#p841486 Hope it helps.

rxs0005 · Answer

thanks B

nice to have u back

rxs0005 · Answer

B

what I have been realizing ( i feel its wrong now ) is that

x ( x - 1 ) < 0 Means

either x< 0 or x - 1 < 0 or x < 1

so x < 0 or x < 1 how did u get 0 < x < 1 ?

Bunuel · Answer

x(x-1)<0 --> roots are x=0 and x=1 --> "<" sign indicates that the solution lies between the roots: 01.gif If it were: x(x-1)>0 --> the same roots: x=0 and x=1 --> ">" sign indicates that the solution lies to the left of the smaller root and to the right of the larger root: x<0 or x>1: 2.gif Follow the links in my previous post for more on this subject. Hope it helps.

Bunuel · Answer

Bumping for review and further discussion.

jlgdr · Answer

Good that you bumped onto this thread Bunuel

Here's my question

On second statement how do we do critical points?

So we have x>x^4

Therefore x^4-x<0

Now then factorizing we have: x^2 (x-1)(x+1) , we only need to take care of (x-1)(x+1) since x^2 is always >=0

But then I get the range -1<x<1 which is not correct given that for instance when x=0 clearly inequality does not hold

Would you mind showing me where I went wrong here?

Thanks!
Cheers
J

KarishmaB · Answer

Note that the inequality is 
 x^4 - x < 0

So on factoring you get 
 x (x^3 - 1) < 0 
 x(x - 1)(x^2 + x + 1) < 0

Now, (x^2 + x + 1) is always positive. It has no real roots. Hence we might as well assume it to be a positive number and ignore it for the sake of getting our transition points. 
Transition points are 0 and 1. Since we want the negative region, the inequality will be negative between 0 and 1.

0 < x < 1

Though, you should try to understand the relation between x, x^2 and x^3 in the ranges < -1, -1 < x < 0, 0 < x <1 and x > 1 so that you don't need to use much algebra for such questions.

OptimusPrepJanielle · Answer

Is x > x^2 ?

This would hold true if x<1

(1) x is greater than x^3 
 x can be a negative integer. For example x = -2. -2^3 = -8. But -2 is not greater than -2^2 = 4. Not sufficient. 
(2) x is greater than x^4  For this to hold true, x would have to be less than 1. sufficient

stonecold · Answer

If the question really wants to know if x lies in the range (0,1)
hence B is sufficient

gmatexam439 · Answer

Hello VeritasPrepKarishma , Its been a while since i studies all such formulaes like (a+1)^3 or a^3-1 or (a+b+c)^2 etc, so should I revisit each and every of these formulaes? Coming back to the question, Can I say that since x>x^4 and since x^4>0 always therefore x MUST be >0 Thus, x>0 and x^4-x<0 => x(x^3-1)<0 => x^3-1<0 => x^3<1 Now, since x >0 then x^3 will be less than 1 only when 0

KarishmaB · Answer

Some important algebraic identities:

(x + y)^2 = x^2 + y^2 + 2xy

(x - y)^2 = x^2 + y^2 - 2xy

x^2 - y^2 = (x + y)*(x - y)

Others, if needed, you can calculate there and then.

(x + y)^3 = (x + y)(x + y)(x + y) = (x^2 + y^2 + 2xy)(x + y) = x^3 + xy^2 + 2x^2y ...

As for this question, yes you can arrive at 0 < x< 1 through that method too. No problem in the reasoning though I would suggest you to look at the other methods discussed above too. They are more generic.

gmatexam439 · Answer

Thank you Karishma

You guyz are doing a very good job. Will look at the other methods described above as well. Regards

Basshead · Answer

Is x > x^2 ? Is x^2 - x < 0 ? Is x(x-1) < 0 ? Is 0 < x < 1 ? (1) x > x^3 x^3 - x < 0 x(x^2-1) < 0 x (x+1)(x-1) < 0 0 < x < 1 and x < - 1 See graph. X^3 - x <0 in two ranges. Insufficient. (2) x > x^4 x^4 - x < 0 x(x^3 - 1) < 0 x(x - 1)(x^2 + x + 1) < 0 x(x - 1) < 0 0 < x < 1 Directly answers our question. Sufficient. Answer is B.

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Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t

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