\(x^3 < 16x\) --> \(x^3-16x<0\) --> \(x(x^2-16)<0\) --> \(x(x+4)(x-4)<0\).

Roots are -4, 0, and 4. This gives us 4 ranges: \(x<-4\), \(-4<x<0\), \(0<x<4\), and \(x>4\). Now, test some extreme value: for example if \(x\) is very large number then the whole expression is positive. Here comes the trick: since in the fourth range, when \(x>4\), the expression is positive, then in third range it'll be negative, in the second positive, and in the first range it'l be negative again: -+-+. Thus, the ranges when the expression is negative are: \(x<-4\) and \(0<x<4\).

Only answer choice D does not include values of x that are not the solutions of given inequality.

Answer: D.

Solving inequalities:
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D is right.

Plugging in is the key here.

Given : x^3 < 16x
Answer should include at least some of the possible solutions for x, but no values that are not solutions

So, even if for one value of x, the given statement fails, we will rule out that option.

A) |x|<4

x < 4 or -x < 4 i.e x > -4.
-4 < x < 4

say x = 0 , x^3 < 16x doesn't satisfy (0 = 0)


B) X <4

say x = 0 . Explained above

C) X>4

say x = 5, x^3 < 16x doesn't satisfy (125 > 80)

D) X<-4

Say x = -5, x^3 < 16x satisfies (-125 < -80)

E) X>0

say x = 1, x^3 < 16x satisfies (1 < 16)
say x = 5, x^3 < 16x doesn't satisfy (125 > 80)

solving inequality -

x^3<16 x
x^3 - 16x < 0
x{x^2 - 16} < 0

This gives us 2 inequalities
{A} x < 0 - this cannot be solved further
{B} x^2 - 16 < 0 (solving it further)

x^2 < 16 - Now solving for x
-4<x<4 - We got the range for X from {B}

thus x can have 2 solutions
Either [a] x <0 Or -4<x<4

We have to find - equations that contain at least some of the possible solutions for x, but no values that are not solutions?

A) |x|<4 -> implies x lie only between 0 and 4 - Which is not correct as per {a} and {b} solved above
B) X <4 -> x can be 3 or -10 or -100 not conclusive
C) X>4 -> x can be 5 or 100 or 400 - Out of range for solutions of X
D) X<-4 -> [b]PERFECT - as it satisfies Only {a} above and not any other value which cannot be a solution for X E) X>0 -> X ranges from 1 to 100 to 1000 - not conclusive as per the question.

Hope it helps.

I mean
can we write x^3<16x as x^3-16x<0 without knowing X is +ve or -ve?

Rgds
Prasannjeet

Yes, we can. I think you are mixing adding/subtracting a value from both sides of the inequality with multiplying/dividing both sides by some value (which we cannot do if the sign of that value is not known).

So, we can safely subtract 16x from both sides to get x^3-16x<0.

Hope it's clear.
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Bunuel - can you explain how you found the roots? I got it down to X(X+4)(X-4) < 0...so x < 0, X < -4, and X < 4...I know this isn't correct but am unsure where my logic is flawed.

Hi All,

This question is written in a "quirky" way - it's asking for an inequality that includes SOME of the solutions to the prompt (but NOT all of the solutions) and NO values that are NOT solutions. Quirky language almost always implies a pattern of some kind - in this case, the pattern can be found by focusing on the ANSWERS....

The prompt gives us X^3 < 16X

We COULD solve this inequality, but we know form the prompt that some of the work won't be useful - we're looking for SOME of the solutions, but not all of the solutions.

From the answers, we know that the numbers 4, 0 and -4 are worth paying attention to.

Let's start with X = 0.....
0^3 is NOT < 16(0)
0 is NOT a solution, so if it appears in an answer, then that answer is WRONG.
Eliminate A and B.

Next, let's try X = 4
4^3 = 64
16(4) = 64
64 is NOT < 64
4 is NOT a solution, so if it appears in an answer, then that answer is WRONG.
Eliminate E.

X = -4 has no effect on either of the remaining answers, so we have to TEST something else...
If....X = 5
5^3 = 125
16(5) = 80
125 is NOT < 80
5 is NOT a solution, so if it appears in an answer, then that answer is WRONG.
Eliminate C.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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x(x)(x) - 16x < 0

x(x(x) - 16) <0

x(x+4)(x-4)<0

We can observe that the value of x from 1 to 3 will work above which eliminates answer option C. x > 4

x = -5 works which will eliminate answer option A and E

x = -4.1 works which eliminates answer option B.

Hence the answer is D.
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Hi Bunuel,

Is there a easy way to find the 4 ranges mentioned above. Can you Pls. help.
I checked the graphical approach for quadratic inequality but could not understand how the above ranges can be deduced.

After going through all the posts on inequalities, I figured out the reasoning. Thanks a lot to Bunuel, Karishma, and gurupreetsingh for excellent posts on inequalities...

x^3 < 16x
i.e. x^3 - 16x < 0
i.e. x(x^2 - 16) < 0
i.e. (x-4)x(x+4) < 0

Case-1: Either (x-4) is Negative and both x and (x+4) are positive
i.e. x<4 and x>0

Case-2: Or (x-4), x and (x+4) are all Negative
i.e. x < -4 Answer: Option D
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we can easily eliminate all but D.

first, suppose x is positive
divide both sides by x
x^2<16
square root both
x<4
but what if x=-5
-2
-2^3 = -8
but 16*-2 = -32, and the equation is not true.

B - out.
C = x=5. 5^3 = 125. 16*5 = 80. so out.
E = x=5. same as in C. so out.

the choice is between A and D.
suppose x=-2.
it satisfies the condition |x|<4.
-2^3 = -8
-2*16 = -32
now -8 is greater than -32. so A is out, and D is the answer.

So we have only two possible ranges where the equation is less than zero.

[1]: x < -4
[2]: 0<x<4

Option D matches our one of the ranges

x^3 < 16x
x(x^2-16) < 0
(x+4)x(x-4) < 0

So, x<-4 or 0<x<4

Only option D satisfies the condition.

Answer D
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\(x^3\) < 16x
Taking 16x to left hand side we get

\(x^3\) - 16x < 0
=> x ( \(x^2\) - 16) < 0
=> x ( \(x^2\) - \(4^2\)) < 0
=> x (x-4)(x+4) < 0 [ Using \(a^2\) - \(b^2\) = (a-b)*(a+b) ]

Using Wave Method to solve the inequality. Watch this video to learn the method



Since the problem is asking for < 0 so we will take the values in the negative range (-) so answer will be

x < -4 and 0 < x < 4

So, Answer will be D

Let's see all the choices

A. |x| < 4 =< -4 < x < 4 => This contains a range which is not in the solution -4 < x <=0
Watch this video to learn about Absolute Value Basics

B. x < 4 => Same as Option A

C. x > 4 => Not a solution

D. x < -4 => Part of the solution

E. x > 0 => Contains x >= 4 which is not in the solution

Watch the Following Video to Master How to Solve Inequality Problems


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