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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
totally confused !!!
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.
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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.
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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.
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.
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.
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.
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.
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.
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.
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...
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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Re: If x^3 < 16x which of the following includes at least some [#permalink]
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14 Mar 2016, 18:48
alchemist009 wrote:
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
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.
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54% (01:32) correct
