Is x^4 - x 0 ? (1) x x^2

Question

Competition Mode Question

Is  x^4 - x > 0  ?

(1)  x < 0 
(2)  x^4 > x^2

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firas92 · Answer

Is  x^4−x>0 ?
Or, is  x^4>x ?

Note than  x  can be greater than  x^4  only in the range  0≤x≤1  so we can have an answer to the question if we know whether  0≤x≤1 .

(1)  x<0

We know that  x  is not in the range  0≤x≤1

so yes,  x^4>x

1 is sufficient

(2)  x^4>x^2

If  x^4>x^2 ,  x  cannot be in the range  -1≤x≤1  which again means that  x  is not in the range  0≤x≤1

so yes,  x^4>x

2 is sufficient

Answer is  (D)

sxenox · Answer

IMO D
x^4 - x
= x(x-1)(x^2 + x + 1)
As complex numbers are out of scope of gmat, we only need to check for x(x-1)>0
ie is x^2 > x?

1) will always satisfy the above
2) as both sides are positive, we can just square root them

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Staphyk · Answer

Is x^4 - x > 0? Yes ?
Or x^4 -x <= 0? No?

(1) x< 0
(-ve)^4 - (-ve) = +ve + +ve = +ve Yes! (Sufficient)

(2) x^4 > x^2
Say x= -2 —> (-2)^4 > (-2)^2
Now (-2)^4 - (-2) >0 yes

Say x= 2 —> (2)^4 > (2)^2
Now (2)^4 - (2)> 0 yes
(Sufficient)

Hit that D

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Leadership · Answer

Is  x^4 - x > 0  ?

(1)  x < 0 
 It means x = (-) negative. You can also consider example such as X = -1, -2
 x^4  will be positive
  x   will be negative but   -x   will be positive
Hence,  x^4 - x > 0  will be always positive

A D /  B C E

(2)  x^4 > x^2 
 It means X is either negative or positive but greater than 1. So, the value of X will increase with increase in its power.
Sufficient.

'D' is the winner.

rajatchopra1994 · Answer

Explanation:

x^4 - x > 0
x*(x^3 - 1^3) > 0
x*(x-1)*(x^2+x+1) > 0

1. x<0
Put any value less than x
let x = -1, Result will be +ve.
Sufficient.

2. x^4>x^2
x^4 - x^2 >0
x^2 * (x^2-1)>0
x^2*(x+1)*(x-1)>0

Sufficient.

IMO-D

freedom128 · Answer

x^4−x>0? ---> x(x-1)(x^2+x+1)>0
The real question (or condition) is whether x<0 or x>1

(1) x<0
Yes, this satisfies the condition above.
SUFFICIENT

(2) x^4>x^2 ---> x^2(x-1)(x+1)>0
So, x<-1 or x>1. This condition also satisfies the condition above.
SUFFICIENT

FINAL ANSWER IS (D)

debjit1990 · Answer

Ans: D

By using both the equation a & b, we can reach the solution.

Archit3110 · Answer

given
x*(x^3-1)>0
#1
x<0
test with -ve integer and fractions we get always yes ; sufficient
#2
x^4>x^2
always true for integer values whether +/-ve
and also yes for x^4−x> 0
sufficient
IMO D
Is x^4−x> 0?

(1) x<0
(2) x4>x2

exc4libur · Answer

x^4−x>0…x^3(x-1)>0 case :If…x>0…then…x-1>0…x>1 case :If…x<0…then…x-1<0…x<1 (1) x<0 sufic x<0, which is less than x<1 (case ). (2) x^4>x^2 sufic x^4-x^2>0…(x^2-x)(x^2+x)>0… >0 >0…x^2(x-1)(x+1)>0 If…x>0… (x-1) >0…then…x-1>0…x>1 If…x<0… (x+1)>0…then…x+1<0…x<-1 Ans (D)

eakabuah · Answer

We are to determine if x^4-x>0.
x^4-x>0
x(x^3-1)>0
range of values x that satisfy the inequality above is: x<0 and x>1.
So basically we are to determine if x falls within the range given above.

Statement 1: x<0
Sufficient since we know that the range of values that satisfy x^4-x>0 includes x<0.

Statement 2: x^4>x^2
x^4>x^2
x^4-x^2>0
x^2(x-1)(x+1)>0
Range of values of x that satisfy the above inequality are: x<-1 and x>1
But x<-1 is a subset of x<0 while x>1 is a range of x values that satisfies x^4-x>0. Hence all values of x that satisfy x^4>x^2 always yield a value of x^4-x > 0.
Statement 2 is also sufficient.

The answer is D.

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CareerGeek · Answer

x^4 − x > 0 --> x(x^3 - 1) > 0 --> x < 0 or x^3 > 1 --> x < 0 or x > 1 E.g: (-1)^4 > -1 or 2^4 > 2 (1) x < 0 --> Sufficient (2) x^4 > x^2 --> x^4 - x^2 > 0 --> x^2(x^2 - 1) > 0 Since x^2 can NEVER be negative, x^2 - 1 > 0 ALWAYS --> x^2 - 1 > 0 --> (x - 1)(x + 1) > 0 --> x < -1 or x > 1 This solution also falls within the solution range of x < 0 or x > 1 --> Sufficient Option D

lacktutor · Answer

Is x^{4}−x>0 ? ( Statement1 ) x<0 x^{4}−x= x(x^{3}-1)>0 x and x^{3}-1 are negative. --> (-)*(-)= (+) Sufficient . ( Statement2 ): x^{4}>x^{2} x^{4} -x^{2}= x^{2} (x^{2}-1)= x^{2} (x-1)(x+1) >0 --> x<-1 and x >1 x(x^{3}-1)>0 x(x-1)(x^{2}+x+1) >0 x<0 and x>1 Always YES Sufficient The answer is D.

ostrick5465 · Answer

x^4 - x = x*(x^3-1) (1) x<0 => x^3 - 1 <0 => x*(x^3-1) > 0 => x^4 - x > 0 => SUFF (2) x^4 > x^2 => x^4 - x^2 >0 => x^2*(x^2-1) > 0 => x^2 - 1 >0 (Because x^2 \geq{0} ) => (x+1)*(x-1) >0 => x>1 or x<-1 Case x<-1 => x^4 - x > 0 (reference (1) is suff above) Case x>1 => x^2 > 0 and x^2-1 >0 => x^4 - x > 0 => (2) Suff => Choice D

unraveled · Answer

Is x^4 − x > 0 ? x(x^3 - 1) > 0 So, Either both x > 0 and x^3 - 1 > 0 i.e. x > 1 OR both x < 0 and x^3 - 1 < 0 i.e. x < 0 (1) x < 0 For x = -1, x^4 - x > 0 = (-1)^4 - (-1) = 2 > 0 SUFFICIENT. (2) x^4 > x^2 x^4 - x^2 > 0 x^2(x^2 - 1) > 0 i.e. x^2 > 1 Hence x > 1 or x < -1 For x = 2, x^4 − x > 0 2^4 − 2 > 0 14 > 0 For x = -2, (-2)^4 − (-2) > 0 18 > 0 SUFFICIENT. Answer D.

Jawad001 · Answer

x^(4 )− x > 0 ? 
Or, x^(4 )>x, this is true for (I) all negative integers and negative fractions
 (II) all positive integers.
 Statement (1) sufficient.
Statement (2) is true for all negative and positive integers but not for real fractions.
Answer:A

chacinluis · Answer

True.

X^4>X^2 iff x>1 or x<-1
Since it's given that x<0
Then it must be the casé that x<-1
Therefore x^4-x is a positive minus a negative, therefore result is positive.
x^4-x>0

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minustark · Answer

The question asks whether x^4 - x > 0 , or x^4 > x. This can be true when x is an integer, or when x is a negative fraction.

1) x < 0. sufficient as x^4 will always be greater than x in this case.

2) x^4 > x^2, sufficient as it means x will be an integer. As we know when x is a fraction (no matter whether it's a positive or negative), its value decreases as power increases.

D is the answer.

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Is x^4 - x > 0 ? (1) x < 0 (2) x^4 > x^2

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