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Is x>0?

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Is x>0?  [#permalink]

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New post 17 Apr 2018, 02:09
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[GMAT math practice question]

Is \(x>0\)?

\(1) x^3+x^2+x = 1\)
\(2) x^2 – 4x - 5 > 0\)

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Is x>0?  [#permalink]

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New post 17 Apr 2018, 03:37
1
MathRevolution wrote:
[GMAT math practice question]

Is \(x>0\)?

\(1) x^3+x^2+x = 1\)
\(2) x^2 – 4x - 5 > 0\)


Statement 1: \(x^3+x^2+x=1\), implies that \(x≠0\), because in that case LHS=0 and RHS=1, which is not possible.

Also if \(-1≤x<0\), then \(x^2>x\) but \(|x|>|x^2|\) which will mean that \(x^3+x^2+x<0\) (for eg. if \(x=-0.9\), then \(x^2=0.81\) and \(x^3=-0.729\), so net sum will be negative), which is not possible. Now we need to check whether \(x<-1\) or \(x>0\)

\(x^3+x^2+x=1\), add \(1\) to both sides of the equation to get \(x^3+x^2+x+1=1+1\)

\(x^2(x+1)+x+1=2=>(x+1)*(x^2+1)=2\), now \(x^2+1>0\), hence \(x+1>0\), because product of two numbers is \(2\) which is positive

Hence \(x>-1\). But as we have seen above that \(x≠0\) and \(x\) cannot be between \(-1\) and \(0\), therefore \(x>0\). Sufficient

Statement 2: \(x^2 – 4x - 5 > 0=>x^2-5x+x-5>0=>(x-5)*(x+1)>0\)

therefore \(x>5\) or \(x<-1\). Hence Insufficient

Option A
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Re: Is x>0?  [#permalink]

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New post 19 Apr 2018, 02:46
1
1
=>
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.

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

Condition 1):
\(x^3+x^2+x = 1\)
\(=> x(x^2+x+1) = 1\)
\(=> x =\frac{1}{(x^2+x+1)}\)
\(=> x = \frac{1}{(x^2+x+1)} > 0\)since \(x^2+x+1 > 0.\)

Thus, condition 1) is sufficient.


Condition 2):
\(x^2 – 4x -5 > 0\)
\(=> (x+1)(x-5) > 0\)
\(=> x < 1 or x > 5\)

Since the solution set of the inequality, \(x > 0\), from the question does not include the solution set of the inequality from condition 2), \(x < 1\) or \(x > 5\), condition 2) is not sufficient.


Therefore, A is the answer.

Answer: A

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: Is x>0?  [#permalink]

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New post 20 Apr 2018, 23:45
MathRevolution wrote:
=>
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.

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

Condition 1):
\(x^3+x^2+x = 1\)
\(=> x(x^2+x+1) = 1\)
\(=> x =\frac{1}{(x^2+x+1)}\)
\(=> x = \frac{1}{(x^2+x+1)} > 0\)since \(x^2+x+1 > 0.\)

Thus, condition 1) is sufficient.





Condition 2):
\(x^2 – 4x -5 > 0\)
\(=> (x+1)(x-5) > 0\)
\(=> x < 1 or x > 5\)

Since the solution set of the inequality, \(x > 0\), from the question does not include the solution set of the inequality from condition 2), \(x < 1\) or \(x > 5\), condition 2) is not sufficient.


Therefore, A is the answer.

Answer: A

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.



i have a doubt here..
As x=-1 satisfies the first equation, then the the condition X>0 is not sufficient...
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Re: Is x>0?  [#permalink]

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New post 20 Apr 2018, 23:58
dewdrops909 wrote:
MathRevolution wrote:
=>
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.

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

Condition 1):
\(x^3+x^2+x = 1\)
\(=> x(x^2+x+1) = 1\)
\(=> x =\frac{1}{(x^2+x+1)}\)
\(=> x = \frac{1}{(x^2+x+1)} > 0\)since \(x^2+x+1 > 0.\)

Thus, condition 1) is sufficient.





Condition 2):
\(x^2 – 4x -5 > 0\)
\(=> (x+1)(x-5) > 0\)
\(=> x < 1 or x > 5\)

Since the solution set of the inequality, \(x > 0\), from the question does not include the solution set of the inequality from condition 2), \(x < 1\) or \(x > 5\), condition 2) is not sufficient.


Therefore, A is the answer.

Answer: A

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.



i have a doubt here..
As x=-1 satisfies the first equation, then the the condition X>0 is not sufficient...


Hi dewdrops909

x=-1 will not satisfy the equation \(x^3+x^2+x=1\) because at x=-1 it will become

-1+1-1=-1 which is not possible.

You may also refer to my solution above to understand the boundary condition.
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Re: Is x>0?  [#permalink]

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New post 21 Apr 2018, 04:47
niks18 wrote:
dewdrops909 wrote:
MathRevolution wrote:
=>
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.

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

Condition 1):
\(x^3+x^2+x = 1\)
\(=> x(x^2+x+1) = 1\)
\(=> x =\frac{1}{(x^2+x+1)}\)
\(=> x = \frac{1}{(x^2+x+1)} > 0\)since \(x^2+x+1 > 0.\)

Thus, condition 1) is sufficient.





Condition 2):
\(x^2 – 4x -5 > 0\)
\(=> (x+1)(x-5) > 0\)
\(=> x < 1 or x > 5\)

Since the solution set of the inequality, \(x > 0\), from the question does not include the solution set of the inequality from condition 2), \(x < 1\) or \(x > 5\), condition 2) is not sufficient.


Therefore, A is the answer.

Answer: A

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.



i have a doubt here..
As x=-1 satisfies the first equation, then the the condition X>0 is not sufficient...


Hi dewdrops909

x=-1 will not satisfy the equation \(x^3+x^2+x=1\) because at x=-1 it will become

-1+1-1=-1 which is not possible.

You may also refer to my solution above to understand the boundary condition.

yeah my mistake...
thanks.
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Is x>0?  [#permalink]

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New post 23 Apr 2018, 02:11
x2+x+1 --> Contains imaginary cube roots of unity so always positive... D < 0 and a> 0
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Re: Is x>0?  [#permalink]

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New post 23 Apr 2018, 05:03
1
MathRevolution wrote:
[GMAT math practice question]

Is \(x>0\)?

\(1) x^3+x^2+x = 1\)
\(2) x^2 – 4x - 5 > 0\)


We can also apply the quadratics here, i.e., knowledge of the shapes of the graph that a quadratic expression will take.

(1) x^3 + x^2 + x = 1 or x(x^2 + x + 1) = 1.
Now lets look at x^2 + x + 1 here. Its a quadratic expression with coefficient of x^2 as positive, so upward facing parabola, and its discriminant is negative, so the parabola will not cross the x axis at all. This means the parabola, or the graph of x^2 + x + 1 will lie entirely above the x axis, thus x^2 + x + 1 will always have positive values only. Knowing this, now we can apply that since x^2 + x + 1 is positive, and product of 'x' and 'x^2 + x + 1' is '1', which is a positive number, it can only happen when x is also positive. So we conclude that x > 0. Sufficient.

(2) x^2 - 4x - 5 > 0 or (x+1)(x-5) > 0. Here x can take positive value say 6, but it can also take negative value say -2. Not sufficient.

Hence A answer
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Re: Is x>0?  [#permalink]

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New post 23 Apr 2018, 15:05
MathRevolution wrote:
[GMAT math practice question]

Is \(x>0\)?

\(1) x^3+x^2+x = 1\)
\(2) x^2 – 4x - 5 > 0\)


\(2) x^2 – 4x - 5 > 0\)

\(2) (x-5)(x+1) > 0\)

Draw number to test ranges

.......Ok.....-1.......Not ok.......5.........Ok.........

\(x\) could be positive or negative

Insufficient

\(1) x^3+x^2+x = 1\)

Let's solve conceptually on number line

Let \(x=0\)......This is invalid as 0\neq{1}.......So x can't be 0

Let \(x\)=-\(\frac{1}{2}\)..... it will yield negative value

Let \(x=-1\) or less.......... it will yield negative value

Clearly x can't be negative value (no need to calculate aggressively as it negative values will make x^2 more lower)

There is only choice left which is x >0

Sufficient

Answer: A
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Re: Is x>0? &nbs [#permalink] 23 Apr 2018, 15:05
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