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

Rgds
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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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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:
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alchemist009
If x^3 < 16x which of the following includes at least some of the possible solutions for x, but no values that are not solutions?

A. |x| < 4
B. x < 4
C. x > 4
D. x < -4
E. x > 0

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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alchemist009
If x^3 < 16x which of the following includes at least some of the possible solutions for x, but no values that are not solutions?

A. |x| < 4
B. x < 4
C. x > 4
D. x < -4
E. x > 0

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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Attachment:
FullSizeRender (3).jpg
FullSizeRender (3).jpg [ 27.6 KiB | Viewed 9958 times ]
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
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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

Attachment:
image-3.jpg
image-3.jpg [ 12.52 KiB | Viewed 6072 times ]

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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We can't just shift the variables to one side without knowing the sign of the variables.

Just put x= -1

after shifting the inequality becomes x^3-16x<0. Put x=-1, it becomes -1+16<0 that is +15<0, which is not true.
Therefore the variables can not be shifted just like that per my understanding.
Please correct me if i am wrong.
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pranavshah7887
We can't just shift the variables to one side without knowing the sign of the variables.

Just put x= -1

after shifting the inequality becomes x^3-16x<0. Put x=-1, it becomes -1+16<0 that is +15<0, which is not true.
Therefore the variables can not be shifted just like that.
Please correct me if i am wrong.
Bunuel

The right way to solve this problem is to use the concept of higer order inequalitiy range.

1. The value x = -1 does not satisfy the inequality x^3 < 16x.

2. When dealing with an inequality, it is always valid to add or subtract the same value from both sides of the inequality.

3. Consider reviewing the theory of inequalities to improve your understanding of their properties:

Hope it helps.
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