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Data Sufficiency : Is x>3?

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Data Sufficiency : Is x>3?

by sydneyuni2014 » Tue Oct 11, 2016 6:49 pm
Hello

I would appreciate if anyone can confirm if I have the right approach

Q: Is x>3?
1. x > 0
2. Under Root of x3 - 9x + 4 > 2

Answer

1. Not sufficient
2.

We square both sides and get x3 - 9x + 4 > 4
Subtract 4 from both sides : x3 - 9x > 0
Take x common x (x2 - 9) > 0
x (x - 3) (x + 3) > 0
x = +3 or - 3
Not Sufficient

Combine : x is positive and x is +3 . Sufficient [spoiler]Ans = C[/spoiler]

Thanks in advance :)
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by fiza gupta » Thu Oct 13, 2016 5:25 am
Q: Is x>3?
1. x > 0
2. Under Root of x3 - 9x + 4 > 2

1) x>0
NOT SUFFICIENT

2) squaring both sides
=x3 - 9x + 4 > 4
=x3 - 9x > 0
=x(x2 - 9) > 0
it can be true if
(i) x>0 and (x2 - 9)>0 or (ii) x<0 and (x2 - 9)<0

(i) x>0 and (x2 - 9)>0
x>0 and |x|>3 ( x>3 and x<-3)
combining we get x>3 - ans to the question will be "yes"

(ii) x<0 and (x2 - 9)<0
x<0 and |x|<3 ( -3<x<3)
combining we get -3<x<0 - ans to the question will be "No"
INSUFFICIENT

combining: x>3 sufficient
because -3<x<0 will not be consider as x<0

so C
Fiza Gupta
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by Jay@ManhattanReview » Thu Feb 02, 2017 11:20 pm
sydneyuni2014 wrote:Hello

I would appreciate if anyone can confirm if I have the right approach

Q: Is x>3?
1. x > 0
2. Under Root of x3 - 9x + 4 > 2

Answer

1. Not sufficient
2.

We square both sides and get x3 - 9x + 4 > 4
Subtract 4 from both sides : x3 - 9x > 0
Take x common x (x2 - 9) > 0
x (x - 3) (x + 3) > 0
x = +3 or - 3
Not Sufficient

Combine : x is positive and x is +3 . Sufficient [spoiler]Ans = C[/spoiler]

Thanks in advance :)
The problem is highlighted in red. You treated the inequality x (x - 3) (x + 3) > 0 as an equality x (x - 3) (x + 3) = 0.

Since we have x (x - 3) (x + 3) greater than 0, x (x - 3) (x + 3) is a positive quantity. If you put x=-3 or x=+3, x (x - 3) (x + 3) would become 0, which is incorrect.

Let's draw the conclusions from x (x - 3) (x + 3) > 0

1. Either all the three terms x, (x - 3), and (x + 3) are greater than 0, OR,

2. Only two terms are less than 0, while the other is greater than 0.

From (1), we see that to make (x-3) > 0, x > 3. Other terms: x, and (x+3) would be positive.

From (2), we see that is we take x as negative, terms x and (x-3) would be negative. Since we want that ONLY two terms must be negative, thus (x+3) must be positive.

Since x is negative, to keep (x+3) positive, 0 > x > -3.

Hope this helps.

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by Jeff@TargetTestPrep » Wed Feb 08, 2017 6:43 pm
sydneyuni2014 wrote:Hello

I would appreciate if anyone can confirm if I have the right approach

Q: Is x>3?
1. x > 0
2. Under Root of x3 - 9x + 4 > 2
We need to determine whether x is greater than 3.

Statement One Alone:

x > 0

Just knowing that x is greater than 0 is not enough information to determine whether x is greater than 3. Statement one alone is not sufficient.

Statement Two Alone:

√(x^3-9x+4) > 2

To get rid of the square root, let's square each side of the inequality:

x^3 - 9x + 4 > 4

Cancelling the 4s on each side, we get:

x^3 - 9x > 0

x(x^2 - 9) > 0

x(x - 3)(x + 3) > 0

We see that x could be greater than 3 or less than 3. If x = -2, then we see that x(x - 3)(x + 3) is greater than 0. However, if x = 4, then x(x - 3)(x + 3) is again greater than 0. In the former case, x < 3, and in the latter case, x > 3. Statement two alone is not sufficient.

Statements One and Two Together:

Using the information from statements one and two, we know that x(x - 3)(x + 3) > 0 and that x > 0. Thus, we can divide each side of the inequality x(x - 3)(x + 3) > 0 by x to obtain:

(x - 3)(x + 3) > 0

Furthermore, since x > 0 and x + 3 > 0, we can divide both sides by x + 3 to obtain:

x - 3 > 0

x > 3

With the two statements, we've determined that x is indeed greater than 3.

Answer: C

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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