Is x^3-4x>0?

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Is x^3-4x>0?

by Max@Math Revolution » Thu Nov 22, 2018 11:40 pm

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[Math Revolution GMAT math practice question]

Is x^3-4x>0?

1) x > 2
2) x > -2

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by fskilnik@GMATH » Fri Nov 23, 2018 3:10 pm

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Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Is x^3-4x>0?

1) x > 2
2) x > -2
$${x^3} - 4x\,\,\mathop > \limits^? \,\,0\,\,\,\,\, \Leftrightarrow \,\,\,\,\boxed{\,x\left( {{x^2} - 4} \right)\,\,\mathop > \limits^? \,\,0\,\,}$$

$$\left( 1 \right)\,\,x > 2\,\,\,\, \Rightarrow \,\,\,\left\{ \matrix{
\,x > 0 \hfill \cr
\,{x^2} > 4\,\,\, \Rightarrow \,\,\,{x^2} - 4 > 0 \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$

$$\left( 2 \right)\,\,x > - 2\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,x = - 1\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,x = \,\,0\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\,\,\,$$

The correct answer is therefore (A).


This solution follows the notations and rationale taught in the GMATH method.

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Fabio.
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by Jay@ManhattanReview » Sat Nov 24, 2018 11:41 pm

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Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Is x^3 - 4x > 0?

1) x > 2
2) x > -2
We have x^3 - 4x > 0 => Is x^3 > 4x?

Note that we cannot cancel x from both the sides since we do not know whether x is positive.

Let's take each statement one by one.

1) x > 2

Note that at x = 2, we have x^3 = 4x => 2^3 = 2^3. As x takes up greater than 2 values, LHS (x^3) increases propotionaly more than the RHS. Say x =3, then x^3 > 4x => 3^3 > 4*3 => 27 > 12. Thus, the answer is yes.

2) x > -2

Note that at x = 3 is one of the values which is in within the range of x > -2. We saw this in Statement 1. The answer is Yes.

However, x = 0 is also with x > - 2 range, for which x^3 > 4x => 0^3 = 4*0 => 0 = 0. The answer is No.

Insufficient.

The correct answer: A

Hope this helps!

-Jay
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by Max@Math Revolution » Sun Nov 25, 2018 5:24 pm

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=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

x^3-4x>0
=> x(x^2-4)>0
=> x(x+2)(x-2)>0
=> -2<x<0 or x > 2

Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider the conditions on their own first.

Condition 1)
In inequality questions, the law "Question is King" tells us that if the solution set of the question includes the solution set of the condition, then the condition is sufficient
Since the solution set of the question, -2<x<0 or x > 2, includes the solution set of condition 1), x > 2, condition 1) is sufficient.

Condition 2)
The solution set of the question, -2<x<0 or x > 2, does not include the solution set of condition 2), x > -2, so condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A

If the original condition includes "1 variable", or "2 variables and 1 equation", or "3 variables and 2 equations" etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.