gmatexam439 wrote:
VeritasPrepKarishma wrote:
jlgdr wrote:
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
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.
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<x<1; since for any real number greater than 1, the cube is greater than the number itself.
Thus i can say x<0<1. Hence sufficient.
Is my reasoning correct?
Regards
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.
_________________
Karishma
Veritas Prep GMAT Instructor
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