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
Data Sufficiency : Is x>3?
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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
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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- Jay@ManhattanReview
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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.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
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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We need to determine whether x is greater than 3.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
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
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