If x^3

Question

If x^3 < 16x, which of the following CANNOT be true? (A) |x| > 4 (B) x > −4 (C) x < −4 (D) x > 4 (E) x < 4

erboi · Accepted Answer

Let's first simplify the expression:
x^3 < 16x
x^3 - 16x <0
x(x-4)(x+4)<0

For the expression above hold true, x<-5 (i.e. -5, -10) or 4>x>0 (i.e. 1,2) -- one can easily visualize this using the number line

Now that we know for which values the expression is true, let's eliminate the alternatives that CAN be true:
(A) |x| > 4 -> true for x=-5
(B) x > −4 -> true for x=1
(C) x < −4 -> true for x=-5
(D) x > 4 -> not true for any of our values
(E) x < 4 -> true for x=1

IMO D
Please give kudos if you like the explanation

Stanindaw · Answer

Posted from my mobile device

BrentGMATPrepNow · Answer

Testing values is a quick and effective approach...

If x³ < 16x, then it COULD be the case that x = -5, since (-5)³ = -125, 16(-5) = -80, and -125 < -80
Since x can equal -5, we can eliminate answer choices A, C and E, since they all state that x CAN equal -5

Similarly, x³ < 16x, then it COULD be the case that x = 1, since 1³ = 1, 16(1) = 16, and 1 < 16
Since x can equal 1, we can eliminate answer choice B since it states that x CAN equal 1

By the process of elimination, the correct answer is D
In other words, x CANNOT be greater than 4.

Manifesting710 · Answer

@ Bunuel where am i going wrong? X^{3} < 16x X^{3} - 16x < 0 x(X^{2} - 4) < 0 x(x +4)(x-4) < 0 , for this to hold true we know x has to be negative using number line + - + ------------(-4)-------------------4--------- so equation will hold true when -4 < x < 4 but arent C & D subset of |x|>4 ? and i understand why x > 4 is the right answer but i dont understand whats wrong in my number line approach

Abhishek009 · Answer

x^3 < 16x Or, x^2 < 16 So, x < + 4 , Thus in a number line the value must lie to the left of +4 inequality-outputter.gif NOw check only (D) points to the right direction , hence mus be incorrect !!!

sherifY · Answer

is the reason i cannot cancle out the X here ( so its X^2<16) that they didnt explicitly say x is not zero?

Bunuel · Answer

No, we cannot reduce x^3 < 16x by x. If x > 0, then you'd keep the sign and get x^2 < 16. If x < 0, then you'd flip the sign and get x^2 > 16. We cannot multiply or divide an inequality by an unknown or expression containing an unknown without knowing its sign. If the unknown is positive, we should preserve the inequality sign, but if it is negative, we should flip the inequality sign. 9. Inequalities Theory Inequalities Made Easy! Inequalities: Tips and hints Arithmetic with Inequalities Wavy Line Method Application - Complex Algebraic Inequalities Solving Quadratic Inequalities: Graphic Approach Inequalities - Quadratic Inequalities Graphic approach to problems with inequalities Inequalities trick INEQUATIONS(Inequalities) Part 1 INEQUATIONS(Inequalities) Part 2 Questions Inequality and absolute value questions from my collection DS Questions PS Questions For more check Ultimate GMAT Quantitative Megathread Hope it helps.

Adarsh_24 · Answer

I think you should consider 3 points -4,0,4. 0- >Since x alone present too. So there are 4 regions. Regions marked in brackets below. --(1)------- -4 ---(2)------ 0 ---(3)-------- 4 --(4)----------- For (1), lets take -5 => -45 < 0 . True For (2), lets take -1 => +15 < 0. False For (3), lets take 1 => -15 <0. True For (4), lets take 5 => 45 <0. False So (2) and (4) are never allowed. Answer x>4 is region (4).

bumpbot · Answer

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If x^3 < 16x, which of the following CANNOT be true?

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