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Bunuel
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

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@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
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Bunuel
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
\(­x^3 < 16x\)

Or, \(x^2 < 16\)

So, \(x < \)­ +\(4\)­, Thus in a number line the value must lie to the left of \(+4\)­
Attachment:
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­ NOw check only (D) points to the right direction , hence mus be incorrect !!!
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is the reason i cannot cancle out the X here ( so its X^2<16) that they didnt explicitly say x is not zero?
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sherifY
is the reason i cannot cancle out the X here ( so its X^2<16) that they didnt explicitly say x is not zero?
­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



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.­
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Manifesting710
@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
­
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).
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